Explanation:
Okay, so I need to figure out how many different values can be obtained from the expression 2·3·4·5 + 1 by inserting parentheses. The original value is 121 because multiplication comes before addition. But if we can insert parentheses, we can get different results. The example given is (2·(3·4))·(5+1) = 144. So my job is to find all possible values by inserting parentheses in different ways and count them.

First, I need to remember that parentheses change the order of operations. Normally, multiplication is done before addition, but inserting parentheses can group operations differently. Also, note that rearranging terms isn't allowed, only inserting parentheses. So the order of the numbers 2, 3, 4, 5, and the +1 must stay the same. The only thing we can do is insert parentheses to change the order in which operations are performed.

The original expression is 2·3·4·5 + 1. Let me write it down with the operations:

2 * 3 * 4 * 5 + 1

So the operations are three multiplications and one addition. The addition is between the product of 2, 3, 4, 5 and 1. But if we can insert parentheses, we can change how these operations are grouped.

First, I need to consider all possible ways to insert parentheses into this expression. However, since there are multiple operations, the number of ways to insert parentheses could be quite large. I need to find a systematic way to approach this.

Let me think about the structure of the expression. There are four numbers and three multiplications, and one addition. The addition is between the product of the first four numbers and 1. So the expression is essentially:

((2 * 3) * 4) * 5 + 1 = 120 + 1 = 121

But if we insert parentheses differently, maybe we can change the order of operations. However, the addition is only between the entire product and 1. Wait, but if we can insert parentheses around parts of the multiplication and the addition, maybe?

Wait, let me think again. The original expression is: 2 * 3 * 4 * 5 + 1. Since multiplication is done before addition, the addition is applied to the result of the multiplications. However, if we can insert parentheses to change the order, we can have different groupings. For example, maybe we can group the addition before some multiplications. But how?

Wait, the addition is between the product of 2,3,4,5 and 1. But perhaps we can insert parentheses such that the addition is done earlier. For instance, if we do (2 * 3 * 4 * (5 + 1)), then that would be different. Wait, but is that allowed?

But in the original expression, the addition symbol is between the product and 1, so inserting parentheses to group 5 + 1 would be allowed. But let me check the original problem statement again: the example given is (2·(3·4))·(5+1). So they grouped the first multiplication as 2*(3*4), then multiplied by (5+1). But in that case, they have parentheses around 3*4 and around 5+1. So in general, parentheses can be inserted to group any operations, as long as the order of numbers is preserved.

Therefore, the key is to consider all possible ways to parenthesize the expression 2 * 3 * 4 * 5 + 1. But how do I systematically do that?

First, let me note that the operations are: three multiplications and one addition. Let me think about where the addition can be. Since addition is a binary operation, it can be placed between two terms. However, in the original expression, the addition is between the product of 2,3,4,5 and 1. But if we want to insert parentheses around parts of the expression, we can have different groupings. But addition can be done before or after some multiplications. However, since the addition is only one, we need to see how it can be grouped with parentheses.

Wait, perhaps I need to think about the expression as having two parts: the multiplicative part and the additive part. But no, the entire expression is a combination of multiplications and additions. Wait, maybe it's helpful to think about the expression as a combination of operations. Let me consider the operations in the expression:

Original expression: 2 * 3 * 4 * 5 + 1

This can be considered as ((2 * 3) * 4) * 5) + 1. But we can also have different groupings:

For example:

- ((2 * 3) * (4 * 5)) + 1
- (2 * (3 * 4)) * 5 + 1
- 2 * ((3 * 4) * 5) + 1
- 2 * (3 * (4 * 5)) + 1
- etc.

But also, we can group the addition with some operations:

For example:

- (2 * 3 * 4 * (5 + 1))
- (2 * (3 * (4 * (5 + 1))))
- ((2 * 3) * (4 * (5 + 1)))
- etc.

So the addition can be grouped with some of the multiplications. The key is to consider all possible ways of inserting parentheses such that the order of operations changes, leading to different results.

But how do I systematically count all these possibilities?

First, perhaps it's helpful to think about the different ways to parenthesize the multiplicative part (the first four numbers) and the additive part. However, since the addition is between the product and 1, inserting parentheses can affect whether the addition occurs before some multiplications or after.

Wait, actually, since there is only one addition, the expression can be considered as having two parts: the product of the numbers 2,3,4,5 and the number 1, connected by addition. However, if we want to change the order, we can group parts of the product with the addition.

So, for example, we can have:

Case 1: The addition is performed first, combining 5 and 1, and then multiplying with the rest.

Case 2: The addition is performed later, after multiplying all the numbers.

But also, there might be intermediate cases where the addition is done at different points. Wait, but the original expression is 2 * 3 * 4 * 5 + 1, so the addition is between the entire product and 1. However, if we insert parentheses around parts of the multiplicative expression, we can affect the order.

Alternatively, perhaps we can consider the expression as two operations: the multiplicative operations and the additive operation. The problem is that addition and multiplication have different precedence, but we are allowed to change that by inserting parentheses.

Therefore, the total expression can be thought of as having two levels of operations: the multiplicative operations (which can be parenthesized in various ways) and the addition. However, the addition can be grouped with different parts of the multiplicative operations.

But this is getting a bit abstract. Maybe I need to model the entire expression as a binary tree. Each binary operation can be represented as a node in a tree. Since we have four numbers and three operations, but the addition is also a binary operation. Wait, the original expression has four numbers and four operations? Wait, no. Let's see:

Wait, the original expression is 2 * 3 * 4 * 5 + 1. That's three multiplications and one addition, so four operations. So, in terms of operations, the expression is:

Starting with numbers 2, 3, 4, 5, and 1. Wait, but hold on, how many numbers do we have? Let me count:

The original expression is 2 * 3 * 4 * 5 + 1. So there are five numbers: 2, 3, 4, 5, and 1. Therefore, there are four operations: three multiplications and one addition. So to parenthesize an expression with five numbers and four operations, the number of different parenthesizations is given by the Catalan number C_4 = 14. However, not all parenthesizations would result in different values because some might be equivalent. But since we have different operations (multiplications and an addition), some of these may produce different values. However, the problem is that the addition is only one, so maybe the total number is less than 14.

But perhaps I need to consider the different ways to parenthesize and see which ones result in different values.

Alternatively, maybe I can think about different ways of grouping the operations, considering that the addition is between two terms, so maybe there are different possibilities based on where the addition is placed in the expression.

Wait, let me think. The original expression is:

2 * 3 * 4 * 5 + 1

So the addition is between the result of 2 * 3 * 4 * 5 and 1. However, if I insert parentheses, I can change when the addition occurs. For example, if I have 2 * 3 * 4 * (5 + 1), then the addition is done first. Similarly, I can have (2 * 3 * 4 * 5) + 1, which is the original.

But also, what if the addition is part of a larger grouping? For example, could we have something like 2 * (3 * (4 * (5 + 1)))? Or (2 * 3) * (4 * (5 + 1))? Or ((2 * 3) * 4) * (5 + 1)?

So, the key is that the addition can be grouped with different parts of the multiplicative operations. Therefore, the different possible values come from different ways of grouping the addition with parts of the multiplicative expression.

Therefore, perhaps I can model this as follows: the addition is between two sub-expressions. One sub-expression is some product involving 2, 3, 4, 5, and the other is 1. Therefore, the different ways to place parentheses correspond to different ways of choosing where to split the multiplicative expression such that the addition is applied to parts of it.

Wait, this seems like a good approach. Let me elaborate. Since the addition is only one operation, and it is between two terms. The first term can be some expression formed by multiplying some of the numbers 2, 3, 4, 5, and the second term is 1. However, since the original expression has 2·3·4·5 + 1, the first term is the entire product. But by inserting parentheses, we can change what is the first term. For example:

- If we do (2·3·4·5) + 1 (original)
- If we do (2·3·4)·(5 + 1)
- (2·3)·(4·(5 + 1))
- 2·((3·4)·(5 + 1))
- 2·(3·(4·(5 + 1)))
- (2·(3·(4·5))) + 1 (Wait, but this is still the same as the original? No, because if we don't group the addition, but the original has the addition at the end. Wait, confusion here.)

Wait, actually, the addition is fixed as the last operation. Wait, no. If we insert parentheses, we can have the addition at different positions. So, for example, the expression could be 2 * 3 * (4 * (5 + 1)), or 2 * (3 * (4 * (5 + 1))), etc. So in each case, the addition is applied to 5 and 1 first, then multiplied with the rest.

Alternatively, the addition could be applied to different parts. But since the original expression only has one addition, the addition is between two terms. Therefore, if we want to insert parentheses, we can choose different ways to split the multiplicative expression into two parts, one of which is combined with 1 via addition.

Therefore, the problem reduces to: how many different ways can we split the multiplicative part (which is 2 * 3 * 4 * 5) into two parts, such that one part is multiplied with (5 + 1) or something else? Wait, maybe not. Let me think again.

If we consider the original expression as (A) + 1, where A is 2 * 3 * 4 * 5. However, if we insert parentheses, we can have different expressions, for example, (B) + 1, where B is a different expression. But B is formed by inserting parentheses into the multiplicative part (2 * 3 * 4 * 5) and combining with 1 via addition. Alternatively, we can have expressions like 2 * (C + 1), where C is some sub-expression. Wait, but where would the addition be? Since the original expression has only one addition, which is between the product and 1. Unless we group some of the multiplicative operations with the addition.

For example, if we want to do something like 2 * (3 * 4 * (5 + 1)), that is, group 5 + 1 first, then multiply with 4, then with 3, then with 2. So in this case, the addition is between 5 and 1, which is grouped with 4, then 3, then 2. Similarly, if we do ((2 * 3) * 4) * (5 + 1). So the addition is only between 5 and 1, and the rest is multiplied.

Alternatively, we could have something like (2 * 3 * (4 * (5 + 1))), which is similar.

But there might be other possibilities where the addition is part of a larger grouping. Wait, but with only one addition, there are limited ways. So perhaps all possible values come from different ways of inserting parentheses that either group the addition with parts of the multiplicative expression or not.

So maybe the number of different values corresponds to the number of different ways to place the addition in the expression, which can be done by choosing different splits of the multiplicative expression into two parts, where one part is combined with 1 via addition and the other part is multiplied with the result.

Wait, let me think. Let's suppose that the addition is between two terms. The first term can be any sub-expression of the multiplicative part (2 * 3 * 4 * 5), and the second term is 1. So, for example:

- The entire multiplicative part: (2 * 3 * 4 * 5) + 1
- The first three numbers multiplied: (2 * 3 * 4) * (5 + 1)
- The first two numbers multiplied: (2 * 3) * (4 * (5 + 1))
- The first number multiplied: 2 * ((3 * 4) * (5 + 1))
- Similarly, maybe also grouping differently?

Wait, but also, could the addition be combined with different operations? For example, 2 * (3 * (4 * (5 + 1))) is the same as grouping the addition with 5 and 1 first, then multiplying with 4, then 3, then 2. Similarly, ((2 * 3) * (4 * (5 + 1))) would be grouping the addition with 5 and 1, multiplying by 4, then by 2 and 3.

Wait, so actually, the different possibilities are determined by where we insert the parentheses that group the addition with different parts of the multiplicative expression. So essentially, the addition is being applied to different "chunks" of the multiplicative expression and then the rest is multiplied accordingly.

But how many different ways are there to split the multiplicative expression into two parts, where one part is multiplied with (5 + 1)? Let me think of the multiplicative expression as 2 * 3 * 4 * 5. To split this into two parts, we can insert a multiplication point between numbers. For example:

- Split after 2: (2) * (3 * 4 * 5)
- Split after 3: (2 * 3) * (4 * 5)
- Split after 4: (2 * 3 * 4) * (5)
- Split after 5: (2 * 3 * 4 * 5) * () but that doesn't make sense...

Wait, but since we want to combine one part with 1 via addition, we need to have the split between the multiplicative part and 1. Wait, perhaps it's better to think that inserting parentheses such that the addition combines some part of the multiplicative expression with the 1.

So for example, the addition can be placed in between different positions. For example, if we consider the multiplicative expression as a sequence, inserting the addition at different positions.

Wait, here's an idea. Since the addition is only between two terms, and we can choose where to put the addition in the expression. So the original expression is:

[2 * 3 * 4 * 5] + 1

But if we want to group the addition at a different place, we can do:

2 * [3 * 4 * 5 + 1]

or

2 * 3 * [4 * 5 + 1]

or

2 * 3 * 4 * [5 + 1]

But also, maybe group it earlier:

[2 * 3] * [4 * 5 + 1]

or

[2 * (3 * 4)] * [5 + 1]

and so on. So the number of different ways depends on how many different positions we can insert the addition into the multiplicative expression.

But how many different positions are there? Let me think. The multiplicative part has four numbers and three multiplications. If we want to insert the addition between different parts, we can think of inserting it after some number of multiplications.

For example:

1. After 2: 2 * (3 * 4 * 5 + 1)
2. After 3: (2 * 3) * (4 * 5 + 1)
3. After 4: (2 * 3 * 4) * (5 + 1)
4. After 5: (2 * 3 * 4 * 5) + 1 (original)

But also, maybe inserting the addition before some multiplications? Wait, but the addition is only once. So we can't have multiple additions. But the previous examples show inserting the addition after different numbers.

But also, the addition could be part of a more nested grouping. For example:

2 * (3 * (4 * (5 + 1)))

Which is equivalent to grouping the addition at the end. So perhaps the different values come from different positions where the addition is inserted into the multiplicative sequence. However, inserting the addition in different positions may lead to different results.

So in total, there are four different positions where the addition can be inserted into the multiplicative sequence, leading to four different values? But in the example given, (2·(3·4))·(5+1) = 144, which is different from the original 121 and from 2*(3*(4*(5+1))) which would be 2*(3*(4*6)) = 2*(3*24)= 2*72=144. Wait, same value? Wait, so maybe some of these are the same. Wait, let me calculate:

Original: 2*3*4*5 +1 = 120 +1=121

Case 1: 2*(3*4*5 +1) = 2*(60 +1)=2*61=122

Case 2: (2*3)*(4*5 +1)= 6*(20 +1)=6*21=126

Case 3: (2*3*4)*(5 +1)= 24*6=144

Case 4: (2*3*4*5)+1=121

But also, 2*(3*(4*(5+1))) = 2*(3*(4*6))=2*(3*24)=2*72=144. So same as case 3. Similarly, ((2*3)*(4*5)) +1= (6*20)+1=121, same as original. Wait, but in that case, if we don't group the addition, is that considered? Wait, no. Wait, the original expression is the addition after the entire product. But if you group differently, but still the addition is after the product. Wait, actually, the original expression is the only case where addition is after the product. So in that case, the other cases are when the addition is inserted earlier. However, if we consider inserting the addition at different depths, perhaps some of them result in same values.

Wait, so in the above, case 1 gives 122, case 2 gives 126, case 3 gives 144, and case 4 gives 121. So that's four different values. However, the example given in the problem was 144, which is case 3. But are there more possibilities?

Wait, maybe if we group addition with different numbers. For example, could we have something like (2 * 3) * (4 * 5 +1) which is case 2, which is 126. What if we have different groupings, like 2 * (3 * 4) * (5 + 1) = (2*12) *6=24*6=144, same as case 3.

Or 2 * ((3 * 4) * (5 + 1)) is same as above.

Alternatively, if we do ((2 * 3) * 4) * (5 + 1) = 24 *6=144.

So all those different parenthesizations lead to same value. So in that case, case 3 is same as other groupings. So the different values are 121, 122, 126, 144. That's four different values. However, the problem says "the expression 2·3·4·5 +1 is equal to 121, since multiplication is carried out before addition. However, we can obtain values other than 121 for this expression if we are allowed to change it by inserting parentheses." So the example given is 144. But according to my previous calculation, there are three other values: 122, 126, 144. So total four values? But maybe there are more?

Wait, but maybe I missed some possibilities. Let me think again.

Is there another way to insert parentheses that would produce different results? For example, can we have the addition happen before multiplying some numbers?

Wait, let's think. Suppose I do 2*(3*(4*5 +1)). Which is 2*(3*(20 +1))= 2*(3*21)=2*63=126. Wait, that's same as case 2. Or 2*((3*4)*5 +1) = 2*(12*5 +1)=2*(60 +1)=2*61=122, same as case 1.

Alternatively, what if we do (2*3*(4*5)) +1 = (2*3*20) +1 = 120 +1=121, same as original.

Alternatively, ((2*3)*4)*5 +1 = same as original.

What about 2*(3*(4*5 +1))? As above, 126.

Or, (2*(3*4)) * (5 +1) = 2*12 *6=144.

What about (2*(3*(4*5))) +1= same as original. So same as 121.

Wait, so apparently, the different ways of inserting parentheses only lead to four different values: 121, 122, 126, 144. So total four?

But let me check if there are more. For example, could we insert parentheses such that the addition is done in between different operations?

Wait, suppose we have ((2 * 3) + 1) * 4 * 5. Is that allowed? But in the original expression, the addition is only between the product of 2,3,4,5 and 1. However, if we insert parentheses to group 2*3 +1, then that would be changing the order of operations. But is that allowed? Wait, the original expression is 2 * 3 * 4 * 5 + 1. If we want to group (2 * 3 + 1) * 4 * 5, that would be inserting parentheses around 2 * 3 + 1. But does that change the original structure?

Wait, the original expression is written as 2 * 3 * 4 * 5 + 1. If we interpret it as (((2 * 3) * 4) * 5) + 1. To group 2 * 3 + 1, we have to have parentheses around 2 * 3 + 1, but then the expression becomes ((2 * 3) + 1) * 4 * 5. Is that considered a valid insertion of parentheses? Let me check.

Original expression: 2 * 3 * 4 * 5 + 1.

If we insert parentheses like (2 * 3 + 1) * 4 * 5, that would require placing parentheses around 2 * 3 + 1. However, the original expression has only one addition. So inserting parentheses around 2 * 3 + 1 is possible. Therefore, that would be a different way of inserting parentheses.

So that would be another value. Let me calculate that:

(2 * 3 + 1) * 4 * 5.

First, 2*3 is 6, 6 +1 is 7. Then 7 * 4 * 5. 7*4=28, 28*5=140. So that's 140. Which is different from the previous ones. So that's another value. Therefore, there are more than four.

Similarly, maybe other groupings?

Let me check. For example, 2 * (3 * 4 * 5 + 1) = 122, which we had.

What if we do 2 * (3 * (4 * 5 + 1))? That's 2*(3*(20 +1))= 2*(3*21)= 126.

Alternatively, 2 * ((3 * 4) * 5 + 1) = 2*(12*5 +1)= 2*(60 +1)=122. Same as first case.

What if we do (2 * 3) * (4 * 5 + 1) = 6 * (20 +1)= 6*21=126.

Or ((2 * 3) + 1) * 4 * 5 = 7 * 20 = 140.

Another possibility: 2 * ( (3 + 1) * 4 * 5 ). Wait, but that would require inserting parentheses around 3 + 1. But the original expression is 2 * 3 * 4 * 5 + 1. If we do 2 * ( (3 + 1) * 4 * 5 ), that would be inserting parentheses around (3 + 1), but that requires changing the original expression by adding parentheses around 3+1, which is allowed? Wait, the original has +1 at the end. If we want to do 3 + 1, that would be inserting parentheses around 3 + 1, which would require changing the original expression, but since the original is 2 * 3 * 4 * 5 + 1, inserting parentheses around 3 + 1 would be (2 * (3 + 1) * 4 * 5). Wait, but is that allowed? Let me check.

Original expression: 2 * 3 * 4 * 5 + 1. If we insert parentheses around 3 + 1, we have to have the addition between 3 and 1. But the original has the addition between the product and 1. Therefore, inserting parentheses around 3 + 1 would require changing the structure of the expression. However, since the original expression has the addition at the end, inserting parentheses to do 3 + 1 would require that the addition is between 3 and 1, which is allowed. Therefore, that would be valid. Let me check:

So, (2 * (3 + 1) * 4 * 5). Wait, but that would be 2 * ((3 +1) * 4 * 5). Let me compute that:

(3 +1)=4, then 4 * 4 * 5 = 16 * 5 = 80, then 2 * 80 = 160. So that would be another value: 160. So that's different from previous ones. So that's another value.

Similarly, could we do other groupings? For example, (2 * 3 * 4) * (5 + 1) = 24 * 6 = 144.

Or (2 * (3 * (4 * (5 + 1)))) = 2 * (3 * (4 * 6)) = 2 * (3 * 24) = 2 * 72 = 144.

Wait, but earlier we had ((2 * 3) +1) * 4 * 5 = 140.

Also, if we group the addition earlier, like ((2 * 3) * 4) + (5 + 1)? Wait, no, because the original expression only has one addition. Wait, no, if we do ((2 * 3) * 4) + (5 + 1), that would be adding two terms: (2*3*4) and (5 +1). But the original expression is only one addition. So inserting two additions would be invalid, unless we group the existing addition with parentheses. Wait, but no, we can't add more additions; we can only insert parentheses. So inserting parentheses can't create new operations. The original expression has only one addition. So inserting parentheses can only group operations, but not add new operations. Therefore, the addition can only be done once, between two terms. Therefore, the expression can be considered as a combination of multiplicative parts and the addition. Therefore, when I considered (2 * (3 + 1) * 4 * 5), that is inserting parentheses to perform 3 +1, which is allowed because the original has that addition. So 3 +1 is part of the original expression. Wait, but the original expression is 2*3*4*5 +1. So the addition is between the product and 1. If we want to group 3 +1, that would mean changing the original expression. Wait, is that allowed?

Wait, here's the confusion. The original expression is 2 * 3 * 4 * 5 + 1. The addition symbol is between the entire product and 1. So in the original expression, the addition is at the end. If we want to group a different addition, say between 3 and 1, that would require that the original expression had that addition. But in the original expression, there's no addition between 3 and 1. Therefore, inserting parentheses to group 3 +1 would require that we change the original expression's structure, which is not allowed. Wait, but maybe not? Wait, the original expression is written as 2 * 3 * 4 * 5 + 1. If we insert parentheses around 3 and 1, we would have to have something like 2 * (3) * (4 * 5) + 1, but that's the same as original. Wait, but if we want to do 2 * (3 + 1) * 4 * 5, then the addition is between 3 and 1, but in the original expression, the addition is between the entire product and 1. Therefore, that would require that the addition is between 3 and 1 instead of between the product and 1. Therefore, that would be changing the order of operations in a way that is not allowed unless parentheses are inserted to enforce that.

But the problem statement says that we can only insert parentheses, not rearrange terms. However, in the original expression, the addition is only at the end. So inserting parentheses to group 3 +1 is possible by inserting parentheses around 3 +1. But wait, how is that possible? Because the original expression is written as 2 * 3 * 4 * 5 + 1. If you insert parentheses around 3 + 1, you would have 2 * (3 + 1) * 4 * 5. However, the original expression has the 5 and 1 as separate terms. To group 3 +1, you would have to have the addition between 3 and 1, which requires that the original expression had the +1 adjacent to 3. But in the original expression, the +1 is at the end. Therefore, to have 3 +1, you need to have the +1 adjacent to 3, which it isn't. Therefore, inserting parentheses to do that would require adding a parentheses that connects 3 and 1, but they aren't adjacent in the original expression. Therefore, that is not allowed.

Wait, this is a critical point. The original expression is 2 * 3 * 4 * 5 + 1. The operations are performed in the order of multiplication before addition. Therefore, the expression is (((2 * 3) * 4) * 5) + 1. However, if we want to insert parentheses to group 3 +1, that would require that 3 and 1 are adjacent in the original expression. However, in the original expression, they are not adjacent. The 3 is followed by 4, and the 1 is at the end. Therefore, to group 3 and 1, we would need to have 3 and 1 adjacent, which they are not in the original expression. Therefore, inserting parentheses to do that is not allowed unless we rearrange the terms. But the problem says that rearranging terms is not allowed, only inserting parentheses. Therefore, such an insertion is not allowed. Therefore, my previous calculation of 160 was incorrect, because it required grouping 3 and 1 which are not adjacent in the original expression. Therefore, that is invalid.

Therefore, we need to be careful about how parentheses can be inserted. The parentheses must not change the order of the numbers. So the numbers must remain in the order 2, 3, 4, 5, 1. Therefore, the addition is only between the product of 2,3,4,5 and 1, but inserting parentheses can only group parts of the multiplicative operations.

Therefore, let me correct that. The addition is between the product of the first four numbers and 1. Therefore, any parentheses that group the addition must be between a sub-expression of the first four numbers and 1. Therefore, the different ways to insert parentheses would correspond to grouping the addition with different parts of the first four numbers.

So, for example, the addition can be between:

- The entire product of 2,3,4,5 and 1: ((2*3*4*5) +1)
- The product of the first three numbers and 5+1: (2*3*4)*(5+1)
- The product of the first two numbers and 4*(5+1): (2*3)*(4*(5+1))
- The first number and (3*4*(5+1)): 2*(3*4*(5+1))
- Or also, if we do 2*(3*(4*(5+1))), which is same as above.

But also, perhaps grouping the addition earlier? For example, ((2*3)*(4*5)) +1, which is same as original. Or ((2*(3*4))*5) +1, which is same as original. So those are same as original.

Therefore, these four cases:

1. ((2*3*4*5) +1) = 121

2. (2*3*4)*(5+1) = 24*6=144

3. (2*3)*(4*(5+1)) = 6*(4*6)=6*24=144

4. 2*(3*4*(5+1)) = 2*(3*4*6)=2*(72)=144

Wait, but all three of these (cases 2, 3, 4) give 144. So that's one unique value. Then case 1 gives 121. So only two values? But the problem says that in the example they obtained 144, but also mentions that we can get other values. Wait, but earlier I thought that inserting parentheses earlier could lead to different values. But according to this, only 121 and 144. But the example given is 144, so maybe there are more.

Wait, maybe I missed some possibilities where the addition is part of a combined operation with more multiplications. Wait, for example:

What if we do (2 * (3 * (4 * (5 + 1)))) which is same as 2*(3*(4*6))=2*3*24=144.

Alternatively, if we do ((2 * 3) * (4 * 5)) + 1=6*20 +1=121, same as original.

Alternatively, what if we do (2 * ((3 * 4) * 5)) + 1= (2 * 12*5) +1= 120 +1=121.

So still same as original.

What about other groupings? For example, can we do something like (2 * 3) * 4 * (5 +1). Which is 6 * 4 *6= 6*24=144. Same as before.

Alternatively, what if we do 2 * ((3 * 4) * 5) +1= same as original.

So, it seems like only two values: 121 and 144. But the problem says that we can obtain other values, like the example given. However, the original problem statement says that the example is 144, but the question is asking for all values obtainable. But according to this, only 121 and 144. However, the original problem says "we can obtain values other than 121", so there must be more than one other value. Therefore, my analysis is missing something.

Wait, going back to the initial problem statement: the example given is (2·(3·4))·(5+1) = 144, which is one of the cases I considered. However, the problem says that there are other values. Let me think again.

Perhaps I need to consider that the addition can be inserted in different parts of the expression, not only with the last number. Let me think again about the expression:

Original expression: 2 * 3 * 4 * 5 + 1

But if I interpret the expression as allowing the addition to be anywhere, as long as the order of numbers is preserved. For example, could we have something like 2 * 3 * (4 + 5) *1? Wait, but the original has +1 at the end. If we insert parentheses to do 4 +5, but then where is the 1?

Wait, the original expression is 2 * 3 * 4 * 5 + 1. If we want to do 4 +5, that would require inserting parentheses around 4+5, and then what about the 1? The original has +1 at the end. If we do 2 * 3 * (4 +5) * 1, that would be changing the original expression by inserting the multiplication with 1. But the original expression has a +1, not a *1. So that would be invalid because we are changing the operation from addition to multiplication. Therefore, that is not allowed unless we alter the expression, which we can't. Therefore, inserting parentheses cannot change the operations. Therefore, the only allowed operations are the existing ones: three multiplications and one addition.

Therefore, the only way to insert parentheses is to group the existing operations differently. Therefore, the addition is between two terms: one being some product of the first four numbers (2*3*4*5), and the other being 1. Therefore, the only way to insert parentheses is to group the addition with different parts of the product.

But then why does the example give 144? Which is achieved by grouping (5 +1) and multiplying with the rest. So that's one possibility. However, the original value is 121. But maybe there is another way?

Wait, let me try to think differently. Suppose we consider that the entire expression is a combination of operations, and we can parenthesize it in any way, as long as the order of numbers is preserved. Therefore, it's similar to fully parenthesizing an expression with four multiplications and one addition.

But how does that work? Let me think of the expression as having the following operations:

Let me represent the expression as a sequence: 2, *, 3, *, 4, *, 5, +, 1.

So there are four operations: three multiplications and one addition. Therefore, inserting parentheses would result in different orders of operations.

To compute all possible values, I need to consider all possible ways to parenthesize this expression. Since there are four operations, the number of different parenthesizations is given by the Catalan number C_4 = 14. But not all of them will give different values.

But since there are both multiplication and addition operations, and multiplication is associative and commutative, but addition is not, so different parenthesizations can lead to different results.

However, given that the numbers are fixed in order, the different parenthesizations would correspond to different orders of applying operations.

So, to find all possible values, I need to consider all possible ways to parenthesize the expression and compute the result for each.

This seems complex, but perhaps manageable. Let's try.

First, let me note that the expression has five numbers: 2, 3, 4, 5, 1 and four operations: three multiplications and one addition. Let me denote the operations as follows:

Positions between numbers:

Between 2 and 3: op1 (multiplication)

Between 3 and 4: op2 (multiplication)

Between 4 and 5: op3 (multiplication)

Between 5 and 1: op4 (addition)

Therefore, the expression is: (op1)(op2)(op3)(op4), with op4 being addition.

The problem is to parenthesize this expression. The number of different parenthesizations is 14, but let me try to find them.

However, it's going to be tedious, but perhaps manageable.

First, note that the operations are op1, op2, op3 (multiplication) and op4 (addition).

To parenthesize this, we can think of the expression as a binary tree with five leaves (the numbers) and four internal nodes (the operations).

The different parenthesizations correspond to different binary trees. However, since there are both multiplications and an addition, some of these trees will result in different values.

Alternatively, since multiplication is associative and commutative, but addition is neither, the different parenthesizations will matter depending on where the addition is.

Since the addition is only one, the key is to determine in which of the operations (op1, op2, op3, op4) the addition is performed, and how the rest are grouped.

Wait, but the addition is fixed as op4. However, by inserting parentheses, we can change the order in which the addition is performed. For example, if we do the addition before some multiplications, or after.

But since the original expression has the addition at the end, op4 is the last operation. However, inserting parentheses can change the order. For example, we can do op4 first, or op1, op2, or op3 first.

Therefore, the addition can be performed at different stages, depending on the parentheses.

Let me think of the possible scenarios where the addition is applied at different points:

Case 1: Addition is applied last (original). So the expression is (((2*3)*4)*5)+1 = 121.

Case 2: Addition is applied before some multiplications. For example, group 5+1 first. So, if we do 5+1 first, then multiply with the rest. Therefore, the expression becomes ((2*3*4)*(5+1)) = 24*6=144.

Case 3: Addition is applied after some multiplications but not all. For example, maybe group 3*4 first, then add 1, etc. Wait, but how?

Wait, for example, suppose we do 2*(3*(4*(5 +1))) as in the example, which results in 144. But that is the same as case 2.

Alternatively, what if the addition is between different operations. For example, suppose we do (2*3)*(4*5) +1 = 6*20 +1= 121, same as original. Or (2*(3*4)) * (5 +1) = 2*12 * 6= 144.

Alternatively, suppose we do 2*(3*(4*5 +1)) = 2*(3*(20 +1))= 2*63=126. Wait, but how is that possible? If we have 4*5 +1, which is 21, then multiplied by 3 (giving 63), then multiplied by 2. But in the original expression, the addition is between the product of 2*3*4*5 and 1. To get 4*5 +1, we need to have the addition between 4*5 and 1, which requires inserting parentheses around 4*5 +1. So, the original expression would need to have the +1 after 4*5. However, in the original expression, the +1 is after the entire product. But if we insert parentheses to do 4*5 +1, does that change the order?

Wait, the original expression is 2 * 3 * 4 * 5 + 1. If we want to group (4 * 5 + 1), that requires that the addition occurs between 4*5 and 1, which is allowed by inserting parentheses. Therefore, the expression becomes 2 * 3 * (4 * 5 + 1).

Yes, that is allowed. Therefore, that is another case: 2 * 3 * (4*5 +1) = 2*3*(20 +1) = 6*21 = 126.

Similarly, if we group (3 * 4 *5 +1), it would be 2*(3*4*5 +1) = 2*(60 +1) = 2*61 = 122.

So, that's another value. So now, we have three values: 121, 122, 126, 144.

Wait, so let's list them:

1. Original: (((2*3)*4)*5) +1 = 121

2. (2*3*4)*(5+1) = 24*6=144

3. 2*(3*(4*(5 +1))) = 2*3*24=144

4. 2*(3*(4*5 +1)) = 2*3*21=126

5. 2*(3*4*(5 +1)) = 2*12*6=144

6. (2*(3*4))* (5+1) = (2*12)*6=24*6=144

7. 2*(3*4*5 +1) = 2*(60 +1)=122

So, the different values so far are 121, 122, 126, 144. So four values. However, are there more?

Let me check if there are other ways to parenthesize where the addition is applied at different points.

For example, if we do ((2*3)* (4 +5)) +1. But wait, does this make sense? Let me see. The original expression is 2 * 3 * 4 * 5 +1. To do (4 +5), we need to have parentheses around 4 and 5. But in the original sequence, the numbers are 2, 3, 4, 5, 1. To group 4 and 5, we need to have something like 2 * 3 * (4 + 5) * 1? No, because the original has +1 at the end. So grouping 4 +5 would require that the entire expression is (...). Wait, let me think.

If I want to do (4 + 5) in the expression, it would have to be part of the operations. But since the original addition is between 5 and 1, to do (4 +5), I would need to have the addition between 4 and 5, but the original has a multiplication between 4 and 5. Therefore, to do (4 +5), we would need to change the operation between 4 and 5 from multiplication to addition, which is not allowed. Therefore, this is invalid. Therefore, we can't do (4 +5). Only the existing addition can be used, which is between 5 and 1.

Therefore, the only place where the addition can be inserted is between 5 and 1, or in a way that groups some multiplicative terms with the +1. Therefore, the other values we have are 121, 122, 126, 144.

But let me check another possibility. What if we do (2 * (3 * 4 * 5)) + 1 = same as original.

Or ((2 * 3) * (4 * 5)) + 1 = same as original.

What about (2 * 3) * (4 * 5) +1? Same as original.

Alternatively, ((2 * (3 * 4)) * 5) +1 = same as original.

So no new values there.

Is there a way to apply the addition before the multiplication with 2? For example:

(2 + 1) * 3 * 4 * 5. Wait, but the original expression has the +1 at the end. To do 2 +1, the addition would have to be between 2 and 1, which are not adjacent in the original expression. Therefore, this would require inserting parentheses to group 2 +1, which is not allowed unless the numbers are adjacent. Hence, invalid.

Similarly, (2 * 3 +1) * 4 * 5: as before, requires grouping 3 +1 which are not adjacent. Therefore, invalid.

Therefore, those are not allowed.

Therefore, the only valid ways to insert parentheses are the ones that involve grouping the existing addition (between 5 and 1) with different parts of the multiplicative expression. So that gives us the four values: 121, 122, 126, 144.

Wait, let me think if there are more ways. What if we have a more nested expression? For example:

2 * (3 * (4 + 5)) * 1? No, because that would require changing the operation between 4 and 5 to addition, which isn't allowed.

Or ((2 * 3) * 4) * (5 +1) = 24 *6 = 144.

Alternatively, (2 * ((3 * 4) * 5)) +1 = same as original.

Is there another way? Suppose we do 2 * 3 * (4 +5) *1. But again, changing the operation between 4 and 5 to addition is not allowed.

Alternatively, maybe something like (2 * 3 * 4 * 5) + 1. Which is original.

Hmm. So maybe those are all. Let me check all possible ways:

The addition can be between:

- The entire multiplicative expression and 1: gives 121.

- The product of the first three numbers and the (5 +1): gives 24 *6=144.

- The product of the first two numbers and (4*(5 +1)): gives 6 * 24=144.

- The first number and (3*4*(5 +1)): gives 2 * (3*4*6)=2*72=144.

- The product of the first three numbers and (the result of the last operation). Wait, but the last operation is the addition. Wait, but if we do 2 * 3 * (4 * (5 +1)), that is the third case.

Wait, but also, if we do 2 * (3 * 4) * (5 +1), which is same as 2 * 12 * 6 = 144.

Alternatively, if we do (2 * 3 * 4) * (5 +1) = same as 24 *6 = 144.

So all those give the same value, 144.

Additionally, if we do 2 * (3 * (4 * 5) +1), which is 2*(3*20 +1)=2*61=122.

And 2 * (3 * (4 *5 +1))= 2*(3*(20 +1))=2*63=126.

Wait, but are these two different? Yes: 122 vs 126. Let me verify:

For 122: 2 * (3 * (4 * 5 +1))

First compute 4 *5 =20, then +1 =21, then *3 =63, then *2 = 126. Wait, but I thought it was 2 * (3 * (4*5) +1). If it's 3*(4*5) +1, that would be 3*20 +1=61, then *2=122. But depending on parentheses.

So if it's 2 * ( (3 * (4 *5)) +1 ), then it's 2*(60 +1)=122. If it's 2 * (3 * (4*5 +1)), then it's 2*(3*(20 +1))=2*63=126.

So these are two different values. So these are two different parenthesizations: one where the addition is inside the innermost parentheses, and one where it is in the outer parentheses.

Therefore, we have:

- 121

- 122

- 126

- 144

Is there another value?

What if we do ((2 * 3) * 4) * (5 +1) = 24 *6=144.

What about (2 * (3 * 4)) * (5 +1)= 24 *6=144.

What about 2 * ((3 * 4) * (5 +1))= same as above.

What about ((2 * (3 * 4)) * 5) +1? Which is (2*12*5) +1= 120 +1=121.

So no new value there.

Alternatively, ((2 * 3) * (4 * 5)) +1 = 6*20 +1=121.

What about (2 * ((3 * 4) * (5 +1))) = same as above.

Is there another way to place the addition? For example, 2 * (3 * (4 +5)) * 1? No, because changing the operation between 4 and 5 is not allowed.

Alternatively, 2 * 3 * (4 +5) *1: same issue.

Therefore, I think those four values are the only ones: 121, 122, 126, 144.

But let me confirm by considering all possible parenthesizations.

The expression has four operations: three multiplications (let's call them M1, M2, M3) and one addition (A). Let's think about the different orders in which these operations can be performed.

There are different ways to order the operations, depending on the parentheses. Since multiplication is associative and addition is not, the order in which the operations are performed matters.

However, since multiplication is commutative and associative, the order of multiplying the numbers doesn't change the result, but the addition is only at the end, so inserting parentheses that change when the addition is performed can give different results.

Let me try enumerating all possible parenthesizations that result in different values.

First, consider the addition (A) being performed:

Option 1: A is performed last. That is, all multiplications are done first, then the addition. This gives 2*3*4*5 +1 = 121.

Option 2: A is performed earlier. Let's think about where A can be applied.

To apply A earlier, it has to be between some terms and the term 1. Since 1 is at the end, the only possible way to apply A earlier is by grouping some part of the multiplicative expression with 1. Therefore, the possible groupings are:

- Group 5 and 1 first: (5 +1) =6. Then multiply with the rest. So 2*3*4*6 = 144.

- Group 4 and 5 with 1: (4*5 +1)=21. Then multiply with the rest: 2*3*21=126.

- Group 3, 4, 5 with 1: (3*4*5 +1)=61. Then multiply with 2: 2*61=122.

- Group 2, 3, 4, 5 with 1: (2*3*4*5 +1)=121.

Wait, but that's the same as option 1. Therefore, these are three different possibilities where the addition is applied earlier: 144, 126, 122, and the original 121. Therefore, four values.

But are there other possibilities where the addition is applied in between the multiplicative operations?

For example, suppose we perform the addition after some multiplications but not all. For instance, suppose we do 2*3 first, then add 1, then multiply with the rest. But this would require inserting the addition between 3 and 1, which are not adjacent in the original expression. Therefore, invalid.

Alternatively, suppose we do 2*3*4*5 +1, but insert parentheses to change the order. However, that is still the same as original.

Alternatively, suppose we do something like 2*(3*4*5 +1) = 122, or 2*3*(4*5 +1)=126, or 2*3*4*(5 +1)=144. So these are the three cases where the addition is applied at different points in the multiplicative sequence.

So, indeed, these are the only possibilities, leading to 121, 122, 126, 144.

Thus, the total number of values is 4.

But wait, the problem says that the example gives 144, and we have 121, 122, 126, and 144. That's four distinct values. Therefore, the answer should be 4? But let me double-check if there are more.

Wait, another thought: if we do ((2 * 3) * 4) * (5 + 1) = (2*3*4)*(5+1)=24*6=144.

Or, if we do (2*(3*4))*(5 +1)= same as above.

What if we do (2*((3*4)*5)) +1= same as original.

Alternatively, 2*( (3*(4*5)) +1 )=2*(60 +1)=122.

Or, 2*(3*(4*5 +1))= 2*(3*21)=126.

Or, 2*((3*4)*5) +1= same as original.

So, yes, only those four values.

Therefore, the total number is 4.

Wait, but I remember that in some cases, different parenthesizations can lead to same values. For example, all the ways to group the addition with 5+1 in different positions give same value (144). Similarly, the other groupings give different values.

So, the four distinct values are 121, 122, 126, 144.

Therefore, the answer is 4. But let me check again if there's another value.

Wait, what if we do ( (2 * 3) + 1 ) * 4 * 5? This would be (6 +1)*20=7*20=140. Is this allowed? The numbers are in order 2, 3, 4, 5, 1. The addition here is between 2 and 3, but they are not adjacent in the original expression. Wait, original expression is 2 * 3 * 4 * 5 + 1. To do 2*3 +1, you have to group (2*3 +1). However, the +1 is at the end, so to have 2*3 +1, you need to have the addition between 3 and 1, but they are not adjacent. Hence, this is not allowed. Therefore, this is invalid.

Similarly, if you try to do 2*(3+1) *4*5, you need to have the +1 between 3 and 1, which are not adjacent. So no.

Therefore, 140 is invalid.

Alternatively, is there any other way?

What if we do ((2*3*4) *5) +1, which is same as original.

Or ((2 * (3 * 4)) * 5) +1, same as original.

Alternatively, ((2 * 3) * (4 * 5)) +1, same as original.

So no new values.

Therefore, yes, only four values: 121, 122, 126, 144.

Wait, but let me check if 122 is indeed a valid value.

Yes: 2 * (3 * (4 * 5) +1). Let me verify that.

Original expression: 2 * 3 * 4 * 5 +1. To compute 2 * (3 * (4 * 5) +1), you first compute 4*5=20, then add 1: 21, then multiply by 3: 63, then multiply by 2: 126. Wait, that gives 126, not 122. Wait, I think I made a mistake earlier.

Wait, earlier I thought that 2 * ( (3 * 4 * 5) +1 ) gives 122. Let's compute this:

First, compute 3 * 4 * 5 = 60, then add 1: 61, then multiply by 2: 122. Yes, that's correct. So that is 122.

But if I do 2 * (3 * (4 * 5 +1)), that's 4*5 +1 =21, 3*21=63, 2*63=126.

So there are two different parenthesizations that give different results: 122 and 126.

So, the four values are:

- 121 (original)

- 122 (2 * (3 * 4 * 5 +1))

- 126 (2 * (3 * (4 * 5 +1)))

- 144 ((2 * 3 * 4) * (5 + 1) etc.)

Therefore, four distinct values.

So the answer should be 4. However, I need to confirm if there are any other ways.

Let me think of another way: for example, (((2 * 3) + 1) * 4) * 5. But this would be adding 2 and 3, but they are not adjacent. Invalid.

Another invalid attempt: ((2 + 3) * 4 * 5) +1. Invalid because + is not between 2 and 3.

Thus, no.

Therefore, I think the answer is 4.

However, I recall that in some similar problems, the number might be higher. Let me think again.

Let me think of all possible parenthesizations:

The expression is 2 * 3 * 4 * 5 + 1. Let's list all the possible parenthesizations, considering the order of operations and where the addition can be.

First, the operations are:

Positions:

1. Between 2 and 3: M1

2. Between 3 and 4: M2

3. Between 4 and 5: M3

4. Between 5 and 1: A

So, total four operations.

The different ways to parenthesize this expression can be represented as different orders of applying operations.

The possible ways are determined by the different ways of choosing which operation to apply last.

For example:

1. Apply addition last: perform all multiplications first, then add 1. Value: 121.

2. Apply addition before some multiplication.

Let's consider the different scenarios where the addition is applied before some multiplications:

Case A: Apply addition at the end (op4).

Case B: Apply addition after op1 (between 2 and 3). But to do that, the addition would have to be between 2 and the rest. But how?

Wait, the addition is only between 5 and 1. To apply addition earlier, it has to be combined with some other operations.

For example:

- Apply op4 (addition) between 5 and 1, then multiply with 4: 4*(5 +1) =24, then multiply with 2*3=6: 6*24=144.

- Apply op4 between 4 and 1: Wait, no, op4 is between 5 and 1. To apply addition between 4 and 5, it would require changing op3 from multiplication to addition, which is not allowed.

Hence, the only way to apply the addition earlier is to apply it at op4 (between 5 and 1), then multiply with the rest.

Therefore, the addition can only be applied at op4, but the rest can be grouped in different ways.

Thus, the different groupings are determined by how the multiplications are grouped before applying the addition.

So, the different groupings are:

- Group all multiplications first: (2*3*4*5)+1=121

- Group the first three numbers, then multiply by (5+1): (2*3*4)*(5+1)=144

- Group the first two numbers, then multiply by (4*(5+1)): (2*3)*(4*(5+1))=6*24=144

- Group the first number, then multiply by (3*4*(5+1)): 2*(3*4*(5+1))=144

- Group the first and second numbers, then multiply by (4*5), then add 1: Wait, but that's the original.

Wait, but there's also the case where the addition is applied after some multiplications, like:

- (2*3)*(4*5) +1 = original

- 2*(3*4)*5 +1 = original

- 2*3*(4*5) +1 = original

So these don't give new values.

However, there's also:

- 2*(3*(4*5 +1)) which is 2*(3*(20+1))= 2*63=126

- 2*( (3*4)*5 +1 ) which is 2*(12*5 +1)=2*(60 +1)=122

So these are two different values.

Therefore, adding those two, we have:

121, 122, 126, 144.

Thus, 4 values.

Therefore, the answer should be 4. But let me check if there's another value from different grouping.

For example, what about ((2 * 3) * (4 + 5)) + 1. But as discussed earlier, this would require changing op3 from multiplication to addition, which is not allowed.

Alternatively, ((2 * (3 * (4 * 5))) +1 ) = (2 * 60) +1=121. Same as original.

So, no new value.

Thus, the total is four distinct values: 121, 122, 126, 144.

Therefore, the answer is 4.

But let me verify by considering that the problem says "inserting parentheses" and that the example gives 144. The question is from a competition, and sometimes these problems have more answers, so I want to be sure.

Wait, let's think about all possible values:

1. 121: original.

2. 122: 2*(3*(4*5) + 1)=122.

3. 126: 2*(3*(4*5 +1))=126.

4. 144: 2*(3*4*(5+1))=144, or (2*3*4)*(5+1)=144, etc.

Is there another value? Let me think.

If we do ((2 * 3) * 4 * 5) + 1 = 121.

Or, 2 * (3 * 4 * 5) +1 = 121? Wait, no, that would be 2*60 +1=121. Wait, but this is same as original? No, the original is 2*3*4*5 +1= 120 +1=121. If I do 2*(3*4*5) +1=2*60 +1=121, same value.

Wait, but if I do 2*(3*4*5) +1, is that a different parenthesization? Or is it same as the original?

In terms of grouping, it's 2*(3*4*5) +1, which is different from ((2*3)*4)*5 +1, but since multiplication is associative, it evaluates to the same value. So, even though the grouping is different, the value is same. Therefore, it doesn't contribute to new value.

Thus, the only additional values are 122, 126, and 144. Hence, total four values.

Therefore, I think the answer is 4.

However, to make sure, let me think of another approach. Since there are four operations, and the addition is one of them, the number of distinct values depends on when the addition is applied.

The addition can be applied either after all multiplications (value 121), or after some multiplications.

When applied after some multiplications, the number of possible ways is equal to the number of ways to split the multiplicative sequence into two parts, one of which is multiplied with the result of (5 +1).

As I did before, there are three such splits (excluding the full multiplication), giving values 122, 126, 144.

Hence, adding the original, total four values.

Yes, that makes sense.

Therefore, I conclude that the number of distinct values is 4.

**Final Answer**
\boxed{4}

To solve the problem, we are given the expression:

$$
2 \cdot 3 \cdot 4 \cdot 5 + 1
$$

We are to determine how many **distinct values** this expression can produce when **parentheses are inserted**, **without rearranging the order of the operands** (i.e., the numbers must remain in the order 2, 3, 4, 5, 1).

---

### Step 1: Understand the Original Expression

The original expression is interpreted as:

$$
(2 \cdot 3) \cdot (4 \cdot 5) + 1 = 120 + 1 = 121
$$

This is the default value when all multiplications are performed before the addition.

We are asked to insert **parentheses** in the expression to alter the **order of operations**, and to count how many **distinct values** can be obtained.

---

### Step 2: Consider the Structure

The expression contains four numbers and four **operations**: three multiplications and **one addition**. The key rule is that **you can only insert parentheses**, so you cannot rearrange the numbers or add new operations.

The addition is **only between 5 and 1**, as the original expression is:

$$
2 \cdot 3 \cdot 4 \cdot 5 + 1
$$

Therefore, **the only legal place for the addition** is between the product of the first four numbers and the final term, 1. But we are allowed to **insert parentheses** to **group parts of the multiplication** with the addition, effectively changing when the addition is performed.

---

### Step 3: Determine All Valid Parentheses Configurations

We will consider different ways to **group** the expression, focusing on where the addition occurs.

Some of the valid expressions (with parentheses) are:

1. **Original (default):**
   $$
   (((2 \cdot 3) \cdot 4) \cdot 5) + 1 = 121
   $$

2. **Group 5 and 1 before multiplying the rest:**
   $$
   (2 \cdot 3 \cdot 4) \cdot (5 + 1) = 24 \cdot 6 = 144
   $$

3. **Group 4 and 5 with 1:**
   $$
   (2 \cdot 3) \cdot (4 \cdot (5 + 1)) = 6 \cdot 24 = 144
   $$

4. **Group 3, 4, and 5 with 1:**
   $$
   2 \cdot (3 \cdot (4 \cdot 5 + 1)) = 2 \cdot (3 \cdot 21) = 2 \cdot 63 = 126
   $$

5. **Group the first three numbers with the result of the addition:**
   $$
   2 \cdot (3 \cdot (4 \cdot 5) + 1) = 2 \cdot (60 + 1) = 2 \cdot 61 = 122
   $$

---

### Step 4: Gather All Distinct Values

From the above configurations, we have computed the following results:

- **121** (original: all multiplications, then addition)
- **122** (2 * (3 * (4 * 5) + 1))
- **126** (2 * (3 * (4 * 5 + 1)))
- **144** (multiple distinct parenthesizations, all yield 144)

Each of these values is **unique** and comes from **different valid insertions of parentheses**, without rearranging operands or adding new operations.

---

### ✅ Final Conclusion

There are **4** distinct values that can be obtained by inserting parentheses in the expression **2·3·4·5 + 1**, considering only valid parenthesizations that preserve the order of operands and apply operations as specified.

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many different values can be obtained from the expression 2·3·4·5 + 1 by inserting parentheses. The original value is 121 because multiplication comes before addition. But if we can insert parentheses, we can get different results. The example given is (2·(3·4))·(5+1) = 144. So my job is to find all possible values by inserting parentheses in different ways and count them.

First, I need to remember that parentheses change the order of operations. Normally, multiplication is done before addition, but inserting parentheses can group operations differently. Also, note that rearranging terms isn't allowed, only inserting parentheses. So the order of the numbers 2, 3, 4, 5, and the +1 must stay the same. The only thing we can do is insert parentheses to change the order in which operations are performed.

The original expression is 2·3·4·5 + 1. Let me write it down with the operations:

2 * 3 * 4 * 5 + 1

So the operations are three multiplications and one addition. The addition is between the product of 2, 3, 4, 5 and 1. But if we can insert parentheses, we can change how these operations are grouped.

First, I need to consider all possible ways to insert parentheses into this expression. However, since there are multiple operations, the number of ways to insert parentheses could be quite large. I need to find a systematic way to approach this.

Let me think about the structure of the expression. There are four numbers and three multiplications, and one addition. The addition is between the product of the first four numbers and 1. So the expression is essentially:

((2 * 3) * 4) * 5 + 1 = 120 + 1 = 121

But if we insert parentheses differently, maybe we can change the order of operations. However, the addition is only between the entire product and 1. Wait, but if we can insert parentheses around parts of the multiplication and the addition, maybe?

Wait, let me think again. The original expression is: 2 * 3 * 4 * 5 + 1. Since multiplication is done before addition, the addition is applied to the result of the multiplications. However, if we can insert parentheses to change the order, we can have different groupings. For example, maybe we can group the addition before some multiplications. But how?

Wait, the addition is between the product of 2,3,4,5 and 1. But perhaps we can insert parentheses such that the addition is done earlier. For instance, if we do (2 * 3 * 4 * (5 + 1)), then that would be different. Wait, but is that allowed?

But in the original expression, the addition symbol is between the product and 1, so inserting parentheses to group 5 + 1 would be allowed. But let me check the original problem statement again: the example given is (2·(3·4))·(5+1). So they grouped the first multiplication as 2*(3*4), then multiplied by (5+1). But in that case, they have parentheses around 3*4 and around 5+1. So in general, parentheses can be inserted to group any operations, as long as the order of numbers is preserved.

Therefore, the key is to consider all possible ways to parenthesize the expression 2 * 3 * 4 * 5 + 1. But how do I systematically do that?

First, let me note that the operations are: three multiplications and one addition. Let me think about where the addition can be. Since addition is a binary operation, it can be placed between two terms. However, in the original expression, the addition is between the product of 2,3,4,5 and 1. But if we want to insert parentheses around parts of the expression, we can have different groupings. But addition can be done before or after some multiplications. However, since the addition is only one, we need to see how it can be grouped with parentheses.

Wait, perhaps I need to think about the expression as having two parts: the multiplicative part and the additive part. But no, the entire expression is a combination of multiplications and additions. Wait, maybe it's helpful to think about the expression as a combination of operations. Let me consider the operations in the expression:

Original expression: 2 * 3 * 4 * 5 + 1

This can be considered as ((2 * 3) * 4) * 5) + 1. But we can also have different groupings:

For example:

- ((2 * 3) * (4 * 5)) + 1
- (2 * (3 * 4)) * 5 + 1
- 2 * ((3 * 4) * 5) + 1
- 2 * (3 * (4 * 5)) + 1
- etc.

But also, we can group the addition with some operations:

For example:

- (2 * 3 * 4 * (5 + 1))
- (2 * (3 * (4 * (5 + 1))))
- ((2 * 3) * (4 * (5 + 1)))
- etc.

So the addition can be grouped with some of the multiplications. The key is to consider all possible ways of inserting parentheses such that the order of operations changes, leading to different results.

But how do I systematically count all these possibilities?

First, perhaps it's helpful to think about the different ways to parenthesize the multiplicative part (the first four numbers) and the additive part. However, since the addition is between the product and 1, inserting parentheses can affect whether the addition occurs before some multiplications or after.

Wait, actually, since there is only one addition, the expression can be considered as having two parts: the product of the numbers 2,3,4,5 and the number 1, connected by addition. However, if we want to change the order, we can group parts of the product with the addition.

So, for example, we can have:

Case 1: The addition is performed first, combining 5 and 1, and then multiplying with the rest.

Case 2: The addition is performed later, after multiplying all the numbers.

But also, there might be intermediate cases where the addition is done at different points. Wait, but the original expression is 2 * 3 * 4 * 5 + 1, so the addition is between the entire product and 1. However, if we insert parentheses around parts of the multiplicative expression, we can affect the order.

Alternatively, perhaps we can consider the expression as two operations: the multiplicative operations and the additive operation. The problem is that addition and multiplication have different precedence, but we are allowed to change that by inserting parentheses.

Therefore, the total expression can be thought of as having two levels of operations: the multiplicative operations (which can be parenthesized in various ways) and the addition. However, the addition can be grouped with different parts of the multiplicative operations.

But this is getting a bit abstract. Maybe I need to model the entire expression as a binary tree. Each binary operation can be represented as a node in a tree. Since we have four numbers and three operations, but the addition is also a binary operation. Wait, the original expression has four numbers and four operations? Wait, no. Let's see:

Wait, the original expression is 2 * 3 * 4 * 5 + 1. That's three multiplications and one addition, so four operations. So, in terms of operations, the expression is:

Starting with numbers 2, 3, 4, 5, and 1. Wait, but hold on, how many numbers do we have? Let me count:

The original expression is 2 * 3 * 4 * 5 + 1. So there are five numbers: 2, 3, 4, 5, and 1. Therefore, there are four operations: three multiplications and one addition. So to parenthesize an expression with five numbers and four operations, the number of different parenthesizations is given by the Catalan number C_4 = 14. However, not all parenthesizations would result in different values because some might be equivalent. But since we have different operations (multiplications and an addition), some of these may produce different values. However, the problem is that the addition is only one, so maybe the total number is less than 14.

But perhaps I need to consider the different ways to parenthesize and see which ones result in different values.

Alternatively, maybe I can think about different ways of grouping the operations, considering that the addition is between two terms, so maybe there are different possibilities based on where the addition is placed in the expression.

Wait, let me think. The original expression is:

2 * 3 * 4 * 5 + 1

So the addition is between the result of 2 * 3 * 4 * 5 and 1. However, if I insert parentheses, I can change when the addition occurs. For example, if I have 2 * 3 * 4 * (5 + 1), then the addition is done first. Similarly, I can have (2 * 3 * 4 * 5) + 1, which is the original.

But also, what if the addition is part of a larger grouping? For example, could we have something like 2 * (3 * (4 * (5 + 1)))? Or (2 * 3) * (4 * (5 + 1))? Or ((2 * 3) * 4) * (5 + 1)?

So, the key is that the addition can be grouped with different parts of the multiplicative operations. Therefore, the different possible values come from different ways of grouping the addition with parts of the multiplicative expression.

Therefore, perhaps I can model this as follows: the addition is between two sub-expressions. One sub-expression is some product involving 2, 3, 4, 5, and the other is 1. Therefore, the different ways to place parentheses correspond to different ways of choosing where to split the multiplicative expression such that the addition is applied to parts of it.

Wait, this seems like a good approach. Let me elaborate. Since the addition is only one operation, and it is between two terms. The first term can be some expression formed by multiplying some of the numbers 2, 3, 4, 5, and the second term is 1. However, since the original expression has 2·3·4·5 + 1, the first term is the entire product. But by inserting parentheses, we can change what is the first term. For example:

- If we do (2·3·4·5) + 1 (original)
- If we do (2·3·4)·(5 + 1)
- (2·3)·(4·(5 + 1))
- 2·((3·4)·(5 + 1))
- 2·(3·(4·(5 + 1)))
- (2·(3·(4·5))) + 1 (Wait, but this is still the same as the original? No, because if we don't group the addition, but the original has the addition at the end. Wait, confusion here.)

Wait, actually, the addition is fixed as the last operation. Wait, no. If we insert parentheses, we can have the addition at different positions. So, for example, the expression could be 2 * 3 * (4 * (5 + 1)), or 2 * (3 * (4 * (5 + 1))), etc. So in each case, the addition is applied to 5 and 1 first, then multiplied with the rest.

Alternatively, the addition could be applied to different parts. But since the original expression only has one addition, the addition is between two terms. Therefore, if we want to insert parentheses, we can choose different ways to split the multiplicative expression into two parts, one of which is combined with 1 via addition.

Therefore, the problem reduces to: how many different ways can we split the multiplicative part (which is 2 * 3 * 4 * 5) into two parts, such that one part is multiplied with (5 + 1) or something else? Wait, maybe not. Let me think again.

If we consider the original expression as (A) + 1, where A is 2 * 3 * 4 * 5. However, if we insert parentheses, we can have different expressions, for example, (B) + 1, where B is a different expression. But B is formed by inserting parentheses into the multiplicative part (2 * 3 * 4 * 5) and combining with 1 via addition. Alternatively, we can have expressions like 2 * (C + 1), where C is some sub-expression. Wait, but where would the addition be? Since the original expression has only one addition, which is between the product and 1. Unless we group some of the multiplicative operations with the addition.

For example, if we want to do something like 2 * (3 * 4 * (5 + 1)), that is, group 5 + 1 first, then multiply with 4, then with 3, then with 2. So in this case, the addition is between 5 and 1, which is grouped with 4, then 3, then 2. Similarly, if we do ((2 * 3) * 4) * (5 + 1). So the addition is only between 5 and 1, and the rest is multiplied.

Alternatively, we could have something like (2 * 3 * (4 * (5 + 1))), which is similar.

But there might be other possibilities where the addition is part of a larger grouping. Wait, but with only one addition, there are limited ways. So perhaps all possible values come from different ways of inserting parentheses that either group the addition with parts of the multiplicative expression or not.

So maybe the number of different values corresponds to the number of different ways to place the addition in the expression, which can be done by choosing different splits of the multiplicative expression into two parts, where one part is combined with 1 via addition and the other part is multiplied with the result.

Wait, let me think. Let's suppose that the addition is between two terms. The first term can be any sub-expression of the multiplicative part (2 * 3 * 4 * 5), and the second term is 1. So, for example:

- The entire multiplicative part: (2 * 3 * 4 * 5) + 1
- The first three numbers multiplied: (2 * 3 * 4) * (5 + 1)
- The first two numbers multiplied: (2 * 3) * (4 * (5 + 1))
- The first number multiplied: 2 * ((3 * 4) * (5 + 1))
- Similarly, maybe also grouping differently?

Wait, but also, could the addition be combined with different operations? For example, 2 * (3 * (4 * (5 + 1))) is the same as grouping the addition with 5 and 1 first, then multiplying with 4, then 3, then 2. Similarly, ((2 * 3) * (4 * (5 + 1))) would be grouping the addition with 5 and 1, multiplying by 4, then by 2 and 3.

Wait, so actually, the different possibilities are determined by where we insert the parentheses that group the addition with different parts of the multiplicative expression. So essentially, the addition is being applied to different "chunks" of the multiplicative expression and then the rest is multiplied accordingly.

But how many different ways are there to split the multiplicative expression into two parts, where one part is multiplied with (5 + 1)? Let me think of the multiplicative expression as 2 * 3 * 4 * 5. To split this into two parts, we can insert a multiplication point between numbers. For example:

- Split after 2: (2) * (3 * 4 * 5)
- Split after 3: (2 * 3) * (4 * 5)
- Split after 4: (2 * 3 * 4) * (5)
- Split after 5: (2 * 3 * 4 * 5) * () but that doesn't make sense...

Wait, but since we want to combine one part with 1 via addition, we need to have the split between the multiplicative part and 1. Wait, perhaps it's better to think that inserting parentheses such that the addition combines some part of the multiplicative expression with the 1.

So for example, the addition can be placed in between different positions. For example, if we consider the multiplicative expression as a sequence, inserting the addition at different positions.

Wait, here's an idea. Since the addition is only between two terms, and we can choose where to put the addition in the expression. So the original expression is:

[2 * 3 * 4 * 5] + 1

But if we want to group the addition at a different place, we can do:

2 * [3 * 4 * 5 + 1]

or

2 * 3 * [4 * 5 + 1]

or

2 * 3 * 4 * [5 + 1]

But also, maybe group it earlier:

[2 * 3] * [4 * 5 + 1]

or

[2 * (3 * 4)] * [5 + 1]

and so on. So the number of different ways depends on how many different positions we can insert the addition into the multiplicative expression.

But how many different positions are there? Let me think. The multiplicative part has four numbers and three multiplications. If we want to insert the addition between different parts, we can think of inserting it after some number of multiplications.

For example:

1. After 2: 2 * (3 * 4 * 5 + 1)
2. After 3: (2 * 3) * (4 * 5 + 1)
3. After 4: (2 * 3 * 4) * (5 + 1)
4. After 5: (2 * 3 * 4 * 5) + 1 (original)

But also, maybe inserting the addition before some multiplications? Wait, but the addition is only once. So we can't have multiple additions. But the previous examples show inserting the addition after different numbers.

But also, the addition could be part of a more nested grouping. For example:

2 * (3 * (4 * (5 + 1)))

Which is equivalent to grouping the addition at the end. So perhaps the different values come from different positions where the addition is inserted into the multiplicative sequence. However, inserting the addition in different positions may lead to different results.

So in total, there are four different positions where the addition can be inserted into the multiplicative sequence, leading to four different values? But in the example given, (2·(3·4))·(5+1) = 144, which is different from the original 121 and from 2*(3*(4*(5+1))) which would be 2*(3*(4*6)) = 2*(3*24)= 2*72=144. Wait, same value? Wait, so maybe some of these are the same. Wait, let me calculate:

Original: 2*3*4*5 +1 = 120 +1=121

Case 1: 2*(3*4*5 +1) = 2*(60 +1)=2*61=122

Case 2: (2*3)*(4*5 +1)= 6*(20 +1)=6*21=126

Case 3: (2*3*4)*(5 +1)= 24*6=144

Case 4: (2*3*4*5)+1=121

But also, 2*(3*(4*(5+1))) = 2*(3*(4*6))=2*(3*24)=2*72=144. So same as case 3. Similarly, ((2*3)*(4*5)) +1= (6*20)+1=121, same as original. Wait, but in that case, if we don't group the addition, is that considered? Wait, no. Wait, the original expression is the addition after the entire product. But if you group differently, but still the addition is after the product. Wait, actually, the original expression is the only case where addition is after the product. So in that case, the other cases are when the addition is inserted earlier. However, if we consider inserting the addition at different depths, perhaps some of them result in same values.

Wait, so in the above, case 1 gives 122, case 2 gives 126, case 3 gives 144, and case 4 gives 121. So that's four different values. However, the example given in the problem was 144, which is case 3. But are there more possibilities?

Wait, maybe if we group addition with different numbers. For example, could we have something like (2 * 3) * (4 * 5 +1) which is case 2, which is 126. What if we have different groupings, like 2 * (3 * 4) * (5 + 1) = (2*12) *6=24*6=144, same as case 3.

Or 2 * ((3 * 4) * (5 + 1)) is same as above.

Alternatively, if we do ((2 * 3) * 4) * (5 + 1) = 24 *6=144.

So all those different parenthesizations lead to same value. So in that case, case 3 is same as other groupings. So the different values are 121, 122, 126, 144. That's four different values. However, the problem says "the expression 2·3·4·5 +1 is equal to 121, since multiplication is carried out before addition. However, we can obtain values other than 121 for this expression if we are allowed to change it by inserting parentheses." So the example given is 144. But according to my previous calculation, there are three other values: 122, 126, 144. So total four values? But maybe there are more?

Wait, but maybe I missed some possibilities. Let me think again.

Is there another way to insert parentheses that would produce different results? For example, can we have the addition happen before multiplying some numbers?

Wait, let's think. Suppose I do 2*(3*(4*5 +1)). Which is 2*(3*(20 +1))= 2*(3*21)=2*63=126. Wait, that's same as case 2. Or 2*((3*4)*5 +1) = 2*(12*5 +1)=2*(60 +1)=2*61=122, same as case 1.

Alternatively, what if we do (2*3*(4*5)) +1 = (2*3*20) +1 = 120 +1=121, same as original.

Alternatively, ((2*3)*4)*5 +1 = same as original.

What about 2*(3*(4*5 +1))? As above, 126.

Or, (2*(3*4)) * (5 +1) = 2*12 *6=144.

What about (2*(3*(4*5))) +1= same as original. So same as 121.

Wait, so apparently, the different ways of inserting parentheses only lead to four different values: 121, 122, 126, 144. So total four?

But let me check if there are more. For example, could we insert parentheses such that the addition is done in between different operations?

Wait, suppose we have ((2 * 3) + 1) * 4 * 5. Is that allowed? But in the original expression, the addition is only between the product of 2,3,4,5 and 1. However, if we insert parentheses to group 2*3 +1, then that would be changing the order of operations. But is that allowed? Wait, the original expression is 2 * 3 * 4 * 5 + 1. If we want to group (2 * 3 + 1) * 4 * 5, that would be inserting parentheses around 2 * 3 + 1. But does that change the original structure?

Wait, the original expression is written as 2 * 3 * 4 * 5 + 1. If we interpret it as (((2 * 3) * 4) * 5) + 1. To group 2 * 3 + 1, we have to have parentheses around 2 * 3 + 1, but then the expression becomes ((2 * 3) + 1) * 4 * 5. Is that considered a valid insertion of parentheses? Let me check.

Original expression: 2 * 3 * 4 * 5 + 1.

If we insert parentheses like (2 * 3 + 1) * 4 * 5, that would require placing parentheses around 2 * 3 + 1. However, the original expression has only one addition. So inserting parentheses around 2 * 3 + 1 is possible. Therefore, that would be a different way of inserting parentheses.

So that would be another value. Let me calculate that:

(2 * 3 + 1) * 4 * 5.

First, 2*3 is 6, 6 +1 is 7. Then 7 * 4 * 5. 7*4=28, 28*5=140. So that's 140. Which is different from the previous ones. So that's another value. Therefore, there are more than four.

Similarly, maybe other groupings?

Let me check. For example, 2 * (3 * 4 * 5 + 1) = 122, which we had.

What if we do 2 * (3 * (4 * 5 + 1))? That's 2*(3*(20 +1))= 2*(3*21)= 126.

Alternatively, 2 * ((3 * 4) * 5 + 1) = 2*(12*5 +1)= 2*(60 +1)=122. Same as first case.

What if we do (2 * 3) * (4 * 5 + 1) = 6 * (20 +1)= 6*21=126.

Or ((2 * 3) + 1) * 4 * 5 = 7 * 20 = 140.

Another possibility: 2 * ( (3 + 1) * 4 * 5 ). Wait, but that would require inserting parentheses around 3 + 1. But the original expression is 2 * 3 * 4 * 5 + 1. If we do 2 * ( (3 + 1) * 4 * 5 ), that would be inserting parentheses around (3 + 1), but that requires changing the original expression by adding parentheses around 3+1, which is allowed? Wait, the original has +1 at the end. If we want to do 3 + 1, that would be inserting parentheses around 3 + 1, which would require changing the original expression, but since the original is 2 * 3 * 4 * 5 + 1, inserting parentheses around 3 + 1 would be (2 * (3 + 1) * 4 * 5). Wait, but is that allowed? Let me check.

Original expression: 2 * 3 * 4 * 5 + 1. If we insert parentheses around 3 + 1, we have to have the addition between 3 and 1. But the original has the addition between the product and 1. Therefore, inserting parentheses around 3 + 1 would require changing the structure of the expression. However, since the original expression has the addition at the end, inserting parentheses to do 3 + 1 would require that the addition is between 3 and 1, which is allowed. Therefore, that would be valid. Let me check:

So, (2 * (3 + 1) * 4 * 5). Wait, but that would be 2 * ((3 +1) * 4 * 5). Let me compute that:

(3 +1)=4, then 4 * 4 * 5 = 16 * 5 = 80, then 2 * 80 = 160. So that would be another value: 160. So that's different from previous ones. So that's another value.

Similarly, could we do other groupings? For example, (2 * 3 * 4) * (5 + 1) = 24 * 6 = 144.

Or (2 * (3 * (4 * (5 + 1)))) = 2 * (3 * (4 * 6)) = 2 * (3 * 24) = 2 * 72 = 144.

Wait, but earlier we had ((2 * 3) +1) * 4 * 5 = 140.

Also, if we group the addition earlier, like ((2 * 3) * 4) + (5 + 1)? Wait, no, because the original expression only has one addition. Wait, no, if we do ((2 * 3) * 4) + (5 + 1), that would be adding two terms: (2*3*4) and (5 +1). But the original expression is only one addition. So inserting two additions would be invalid, unless we group the existing addition with parentheses. Wait, but no, we can't add more additions; we can only insert parentheses. So inserting parentheses can't create new operations. The original expression has only one addition. So inserting parentheses can only group operations, but not add new operations. Therefore, the addition can only be done once, between two terms. Therefore, the expression can be considered as a combination of multiplicative parts and the addition. Therefore, when I considered (2 * (3 + 1) * 4 * 5), that is inserting parentheses to perform 3 +1, which is allowed because the original has that addition. So 3 +1 is part of the original expression. Wait, but the original expression is 2*3*4*5 +1. So the addition is between the product and 1. If we want to group 3 +1, that would mean changing the original expression. Wait, is that allowed?

Wait, here's the confusion. The original expression is 2 * 3 * 4 * 5 + 1. The addition symbol is between the entire product and 1. So in the original expression, the addition is at the end. If we want to group a different addition, say between 3 and 1, that would require that the original expression had that addition. But in the original expression, there's no addition between 3 and 1. Therefore, inserting parentheses to group 3 +1 would require that we change the original expression's structure, which is not allowed. Wait, but maybe not? Wait, the original expression is written as 2 * 3 * 4 * 5 + 1. If we insert parentheses around 3 and 1, we would have to have something like 2 * (3) * (4 * 5) + 1, but that's the same as original. Wait, but if we want to do 2 * (3 + 1) * 4 * 5, then the addition is between 3 and 1, but in the original expression, the addition is between the entire product and 1. Therefore, that would require that the addition is between 3 and 1 instead of between the product and 1. Therefore, that would be changing the order of operations in a way that is not allowed unless parentheses are inserted to enforce that.

But the problem statement says that we can only insert parentheses, not rearrange terms. However, in the original expression, the addition is only at the end. So inserting parentheses to group 3 +1 is possible by inserting parentheses around 3 +1. But wait, how is that possible? Because the original expression is written as 2 * 3 * 4 * 5 + 1. If you insert parentheses around 3 + 1, you would have 2 * (3 + 1) * 4 * 5. However, the original expression has the 5 and 1 as separate terms. To group 3 +1, you would have to have the addition between 3 and 1, which requires that the original expression had the +1 adjacent to 3. But in the original expression, the +1 is at the end. Therefore, to have 3 +1, you need to have the +1 adjacent to 3, which it isn't. Therefore, inserting parentheses to do that would require adding a parentheses that connects 3 and 1, but they aren't adjacent in the original expression. Therefore, that is not allowed.

Wait, this is a critical point. The original expression is 2 * 3 * 4 * 5 + 1. The operations are performed in the order of multiplication before addition. Therefore, the expression is (((2 * 3) * 4) * 5) + 1. However, if we want to insert parentheses to group 3 +1, that would require that 3 and 1 are adjacent in the original expression. However, in the original expression, they are not adjacent. The 3 is followed by 4, and the 1 is at the end. Therefore, to group 3 and 1, we would need to have 3 and 1 adjacent, which they are not in the original expression. Therefore, inserting parentheses to do that is not allowed unless we rearrange the terms. But the problem says that rearranging terms is not allowed, only inserting parentheses. Therefore, such an insertion is not allowed. Therefore, my previous calculation of 160 was incorrect, because it required grouping 3 and 1 which are not adjacent in the original expression. Therefore, that is invalid.

Therefore, we need to be careful about how parentheses can be inserted. The parentheses must not change the order of the numbers. So the numbers must remain in the order 2, 3, 4, 5, 1. Therefore, the addition is only between the product of 2,3,4,5 and 1, but inserting parentheses can only group parts of the multiplicative operations.

Therefore, let me correct that. The addition is between the product of the first four numbers and 1. Therefore, any parentheses that group the addition must be between a sub-expression of the first four numbers and 1. Therefore, the different ways to insert parentheses would correspond to grouping the addition with different parts of the first four numbers.

So, for example, the addition can be between:

- The entire product of 2,3,4,5 and 1: ((2*3*4*5) +1)
- The product of the first three numbers and 5+1: (2*3*4)*(5+1)
- The product of the first two numbers and 4*(5+1): (2*3)*(4*(5+1))
- The first number and (3*4*(5+1)): 2*(3*4*(5+1))
- Or also, if we do 2*(3*(4*(5+1))), which is same as above.

But also, perhaps grouping the addition earlier? For example, ((2*3)*(4*5)) +1, which is same as original. Or ((2*(3*4))*5) +1, which is same as original. So those are same as original.

Therefore, these four cases:

1. ((2*3*4*5) +1) = 121

2. (2*3*4)*(5+1) = 24*6=144

3. (2*3)*(4*(5+1)) = 6*(4*6)=6*24=144

4. 2*(3*4*(5+1)) = 2*(3*4*6)=2*(72)=144

Wait, but all three of these (cases 2, 3, 4) give 144. So that's one unique value. Then case 1 gives 121. So only two values? But the problem says that in the example they obtained 144, but also mentions that we can get other values. Wait, but earlier I thought that inserting parentheses earlier could lead to different values. But according to this, only 121 and 144. But the example given is 144, so maybe there are more.

Wait, maybe I missed some possibilities where the addition is part of a combined operation with more multiplications. Wait, for example:

What if we do (2 * (3 * (4 * (5 + 1)))) which is same as 2*(3*(4*6))=2*3*24=144.

Alternatively, if we do ((2 * 3) * (4 * 5)) + 1=6*20 +1=121, same as original.

Alternatively, what if we do (2 * ((3 * 4) * 5)) + 1= (2 * 12*5) +1= 120 +1=121.

So still same as original.

What about other groupings? For example, can we do something like (2 * 3) * 4 * (5 +1). Which is 6 * 4 *6= 6*24=144. Same as before.

Alternatively, what if we do 2 * ((3 * 4) * 5) +1= same as original.

So, it seems like only two values: 121 and 144. But the problem says that we can obtain other values, like the example given. However, the original problem statement says that the example is 144, but the question is asking for all values obtainable. But according to this, only 121 and 144. However, the original problem says "we can obtain values other than 121", so there must be more than one other value. Therefore, my analysis is missing something.

Wait, going back to the initial problem statement: the example given is (2·(3·4))·(5+1) = 144, which is one of the cases I considered. However, the problem says that there are other values. Let me think again.

Perhaps I need to consider that the addition can be inserted in different parts of the expression, not only with the last number. Let me think again about the expression:

Original expression: 2 * 3 * 4 * 5 + 1

But if I interpret the expression as allowing the addition to be anywhere, as long as the order of numbers is preserved. For example, could we have something like 2 * 3 * (4 + 5) *1? Wait, but the original has +1 at the end. If we insert parentheses to do 4 +5, but then where is the 1?

Wait, the original expression is 2 * 3 * 4 * 5 + 1. If we want to do 4 +5, that would require inserting parentheses around 4+5, and then what about the 1? The original has +1 at the end. If we do 2 * 3 * (4 +5) * 1, that would be changing the original expression by inserting the multiplication with 1. But the original expression has a +1, not a *1. So that would be invalid because we are changing the operation from addition to multiplication. Therefore, that is not allowed unless we alter the expression, which we can't. Therefore, inserting parentheses cannot change the operations. Therefore, the only allowed operations are the existing ones: three multiplications and one addition.

Therefore, the only way to insert parentheses is to group the existing operations differently. Therefore, the addition is between two terms: one being some product of the first four numbers (2*3*4*5), and the other being 1. Therefore, the only way to insert parentheses is to group the addition with different parts of the product.

But then why does the example give 144? Which is achieved by grouping (5 +1) and multiplying with the rest. So that's one possibility. However, the original value is 121. But maybe there is another way?

Wait, let me try to think differently. Suppose we consider that the entire expression is a combination of operations, and we can parenthesize it in any way, as long as the order of numbers is preserved. Therefore, it's similar to fully parenthesizing an expression with four multiplications and one addition.

But how does that work? Let me think of the expression as having the following operations:

Let me represent the expression as a sequence: 2, *, 3, *, 4, *, 5, +, 1.

So there are four operations: three multiplications and one addition. Therefore, inserting parentheses would result in different orders of operations.

To compute all possible values, I need to consider all possible ways to parenthesize this expression. Since there are four operations, the number of different parenthesizations is given by the Catalan number C_4 = 14. But not all of them will give different values.

But since there are both multiplication and addition operations, and multiplication is associative and commutative, but addition is not, so different parenthesizations can lead to different results.

However, given that the numbers are fixed in order, the different parenthesizations would correspond to different orders of applying operations.

So, to find all possible values, I need to consider all possible ways to parenthesize the expression and compute the result for each.

This seems complex, but perhaps manageable. Let's try.

First, let me note that the expression has five numbers: 2, 3, 4, 5, 1 and four operations: three multiplications and one addition. Let me denote the operations as follows:

Positions between numbers:

Between 2 and 3: op1 (multiplication)

Between 3 and 4: op2 (multiplication)

Between 4 and 5: op3 (multiplication)

Between 5 and 1: op4 (addition)

Therefore, the expression is: (op1)(op2)(op3)(op4), with op4 being addition.

The problem is to parenthesize this expression. The number of different parenthesizations is 14, but let me try to find them.

However, it's going to be tedious, but perhaps manageable.

First, note that the operations are op1, op2, op3 (multiplication) and op4 (addition).

To parenthesize this, we can think of the expression as a binary tree with five leaves (the numbers) and four internal nodes (the operations).

The different parenthesizations correspond to different binary trees. However, since there are both multiplications and an addition, some of these trees will result in different values.

Alternatively, since multiplication is associative and commutative, but addition is neither, the different parenthesizations will matter depending on where the addition is.

Since the addition is only one, the key is to determine in which of the operations (op1, op2, op3, op4) the addition is performed, and how the rest are grouped.

Wait, but the addition is fixed as op4. However, by inserting parentheses, we can change the order in which the addition is performed. For example, if we do the addition before some multiplications, or after.

But since the original expression has the addition at the end, op4 is the last operation. However, inserting parentheses can change the order. For example, we can do op4 first, or op1, op2, or op3 first.

Therefore, the addition can be performed at different stages, depending on the parentheses.

Let me think of the possible scenarios where the addition is applied at different points:

Case 1: Addition is applied last (original). So the expression is (((2*3)*4)*5)+1 = 121.

Case 2: Addition is applied before some multiplications. For example, group 5+1 first. So, if we do 5+1 first, then multiply with the rest. Therefore, the expression becomes ((2*3*4)*(5+1)) = 24*6=144.

Case 3: Addition is applied after some multiplications but not all. For example, maybe group 3*4 first, then add 1, etc. Wait, but how?

Wait, for example, suppose we do 2*(3*(4*(5 +1))) as in the example, which results in 144. But that is the same as case 2.

Alternatively, what if the addition is between different operations. For example, suppose we do (2*3)*(4*5) +1 = 6*20 +1= 121, same as original. Or (2*(3*4)) * (5 +1) = 2*12 * 6= 144.

Alternatively, suppose we do 2*(3*(4*5 +1)) = 2*(3*(20 +1))= 2*63=126. Wait, but how is that possible? If we have 4*5 +1, which is 21, then multiplied by 3 (giving 63), then multiplied by 2. But in the original expression, the addition is between the product of 2*3*4*5 and 1. To get 4*5 +1, we need to have the addition between 4*5 and 1, which requires inserting parentheses around 4*5 +1. So, the original expression would need to have the +1 after 4*5. However, in the original expression, the +1 is after the entire product. But if we insert parentheses to do 4*5 +1, does that change the order?

Wait, the original expression is 2 * 3 * 4 * 5 + 1. If we want to group (4 * 5 + 1), that requires that the addition occurs between 4*5 and 1, which is allowed by inserting parentheses. Therefore, the expression becomes 2 * 3 * (4 * 5 + 1).

Yes, that is allowed. Therefore, that is another case: 2 * 3 * (4*5 +1) = 2*3*(20 +1) = 6*21 = 126.

Similarly, if we group (3 * 4 *5 +1), it would be 2*(3*4*5 +1) = 2*(60 +1) = 2*61 = 122.

So, that's another value. So now, we have three values: 121, 122, 126, 144.

Wait, so let's list them:

1. Original: (((2*3)*4)*5) +1 = 121

2. (2*3*4)*(5+1) = 24*6=144

3. 2*(3*(4*(5 +1))) = 2*3*24=144

4. 2*(3*(4*5 +1)) = 2*3*21=126

5. 2*(3*4*(5 +1)) = 2*12*6=144

6. (2*(3*4))* (5+1) = (2*12)*6=24*6=144

7. 2*(3*4*5 +1) = 2*(60 +1)=122

So, the different values so far are 121, 122, 126, 144. So four values. However, are there more?

Let me check if there are other ways to parenthesize where the addition is applied at different points.

For example, if we do ((2*3)* (4 +5)) +1. But wait, does this make sense? Let me see. The original expression is 2 * 3 * 4 * 5 +1. To do (4 +5), we need to have parentheses around 4 and 5. But in the original sequence, the numbers are 2, 3, 4, 5, 1. To group 4 and 5, we need to have something like 2 * 3 * (4 + 5) * 1? No, because the original has +1 at the end. So grouping 4 +5 would require that the entire expression is (...). Wait, let me think.

If I want to do (4 + 5) in the expression, it would have to be part of the operations. But since the original addition is between 5 and 1, to do (4 +5), I would need to have the addition between 4 and 5, but the original has a multiplication between 4 and 5. Therefore, to do (4 +5), we would need to change the operation between 4 and 5 from multiplication to addition, which is not allowed. Therefore, this is invalid. Therefore, we can't do (4 +5). Only the existing addition can be used, which is between 5 and 1.

Therefore, the only place where the addition can be inserted is between 5 and 1, or in a way that groups some multiplicative terms with the +1. Therefore, the other values we have are 121, 122, 126, 144.

But let me check another possibility. What if we do (2 * (3 * 4 * 5)) + 1 = same as original.

Or ((2 * 3) * (4 * 5)) + 1 = same as original.

What about (2 * 3) * (4 * 5) +1? Same as original.

Alternatively, ((2 * (3 * 4)) * 5) +1 = same as original.

So no new values there.

Is there a way to apply the addition before the multiplication with 2? For example:

(2 + 1) * 3 * 4 * 5. Wait, but the original expression has the +1 at the end. To do 2 +1, the addition would have to be between 2 and 1, which are not adjacent in the original expression. Therefore, this would require inserting parentheses to group 2 +1, which is not allowed unless the numbers are adjacent. Hence, invalid.

Similarly, (2 * 3 +1) * 4 * 5: as before, requires grouping 3 +1 which are not adjacent. Therefore, invalid.

Therefore, those are not allowed.

Therefore, the only valid ways to insert parentheses are the ones that involve grouping the existing addition (between 5 and 1) with different parts of the multiplicative expression. So that gives us the four values: 121, 122, 126, 144.

Wait, let me think if there are more ways. What if we have a more nested expression? For example:

2 * (3 * (4 + 5)) * 1? No, because that would require changing the operation between 4 and 5 to addition, which isn't allowed.

Or ((2 * 3) * 4) * (5 +1) = 24 *6 = 144.

Alternatively, (2 * ((3 * 4) * 5)) +1 = same as original.

Is there another way? Suppose we do 2 * 3 * (4 +5) *1. But again, changing the operation between 4 and 5 to addition is not allowed.

Alternatively, maybe something like (2 * 3 * 4 * 5) + 1. Which is original.

Hmm. So maybe those are all. Let me check all possible ways:

The addition can be between:

- The entire multiplicative expression and 1: gives 121.

- The product of the first three numbers and the (5 +1): gives 24 *6=144.

- The product of the first two numbers and (4*(5 +1)): gives 6 * 24=144.

- The first number and (3*4*(5 +1)): gives 2 * (3*4*6)=2*72=144.

- The product of the first three numbers and (the result of the last operation). Wait, but the last operation is the addition. Wait, but if we do 2 * 3 * (4 * (5 +1)), that is the third case.

Wait, but also, if we do 2 * (3 * 4) * (5 +1), which is same as 2 * 12 * 6 = 144.

Alternatively, if we do (2 * 3 * 4) * (5 +1) = same as 24 *6 = 144.

So all those give the same value, 144.

Additionally, if we do 2 * (3 * (4 * 5) +1), which is 2*(3*20 +1)=2*61=122.

And 2 * (3 * (4 *5 +1))= 2*(3*(20 +1))=2*63=126.

Wait, but are these two different? Yes: 122 vs 126. Let me verify:

For 122: 2 * (3 * (4 * 5 +1))

First compute 4 *5 =20, then +1 =21, then *3 =63, then *2 = 126. Wait, but I thought it was 2 * (3 * (4*5) +1). If it's 3*(4*5) +1, that would be 3*20 +1=61, then *2=122. But depending on parentheses.

So if it's 2 * ( (3 * (4 *5)) +1 ), then it's 2*(60 +1)=122. If it's 2 * (3 * (4*5 +1)), then it's 2*(3*(20 +1))=2*63=126.

So these are two different values. So these are two different parenthesizations: one where the addition is inside the innermost parentheses, and one where it is in the outer parentheses.

Therefore, we have:

- 121

- 122

- 126

- 144

Is there another value?

What if we do ((2 * 3) * 4) * (5 +1) = 24 *6=144.

What about (2 * (3 * 4)) * (5 +1)= 24 *6=144.

What about 2 * ((3 * 4) * (5 +1))= same as above.

What about ((2 * (3 * 4)) * 5) +1? Which is (2*12*5) +1= 120 +1=121.

So no new value there.

Alternatively, ((2 * 3) * (4 * 5)) +1 = 6*20 +1=121.

What about (2 * ((3 * 4) * (5 +1))) = same as above.

Is there another way to place the addition? For example, 2 * (3 * (4 +5)) * 1? No, because changing the operation between 4 and 5 is not allowed.

Alternatively, 2 * 3 * (4 +5) *1: same issue.

Therefore, I think those four values are the only ones: 121, 122, 126, 144.

But let me confirm by considering all possible parenthesizations.

The expression has four operations: three multiplications (let's call them M1, M2, M3) and one addition (A). Let's think about the different orders in which these operations can be performed.

There are different ways to order the operations, depending on the parentheses. Since multiplication is associative and addition is not, the order in which the operations are performed matters.

However, since multiplication is commutative and associative, the order of multiplying the numbers doesn't change the result, but the addition is only at the end, so inserting parentheses that change when the addition is performed can give different results.

Let me try enumerating all possible parenthesizations that result in different values.

First, consider the addition (A) being performed:

Option 1: A is performed last. That is, all multiplications are done first, then the addition. This gives 2*3*4*5 +1 = 121.

Option 2: A is performed earlier. Let's think about where A can be applied.

To apply A earlier, it has to be between some terms and the term 1. Since 1 is at the end, the only possible way to apply A earlier is by grouping some part of the multiplicative expression with 1. Therefore, the possible groupings are:

- Group 5 and 1 first: (5 +1) =6. Then multiply with the rest. So 2*3*4*6 = 144.

- Group 4 and 5 with 1: (4*5 +1)=21. Then multiply with the rest: 2*3*21=126.

- Group 3, 4, 5 with 1: (3*4*5 +1)=61. Then multiply with 2: 2*61=122.

- Group 2, 3, 4, 5 with 1: (2*3*4*5 +1)=121.

Wait, but that's the same as option 1. Therefore, these are three different possibilities where the addition is applied earlier: 144, 126, 122, and the original 121. Therefore, four values.

But are there other possibilities where the addition is applied in between the multiplicative operations?

For example, suppose we perform the addition after some multiplications but not all. For instance, suppose we do 2*3 first, then add 1, then multiply with the rest. But this would require inserting the addition between 3 and 1, which are not adjacent in the original expression. Therefore, invalid.

Alternatively, suppose we do 2*3*4*5 +1, but insert parentheses to change the order. However, that is still the same as original.

Alternatively, suppose we do something like 2*(3*4*5 +1) = 122, or 2*3*(4*5 +1)=126, or 2*3*4*(5 +1)=144. So these are the three cases where the addition is applied at different points in the multiplicative sequence.

So, indeed, these are the only possibilities, leading to 121, 122, 126, 144.

Thus, the total number of values is 4.

But wait, the problem says that the example gives 144, and we have 121, 122, 126, and 144. That's four distinct values. Therefore, the answer should be 4? But let me double-check if there are more.

Wait, another thought: if we do ((2 * 3) * 4) * (5 + 1) = (2*3*4)*(5+1)=24*6=144.

Or, if we do (2*(3*4))*(5 +1)= same as above.

What if we do (2*((3*4)*5)) +1= same as original.

Alternatively, 2*( (3*(4*5)) +1 )=2*(60 +1)=122.

Or, 2*(3*(4*5 +1))= 2*(3*21)=126.

Or, 2*((3*4)*5) +1= same as original.

So, yes, only those four values.

Therefore, the total number is 4.

Wait, but I remember that in some cases, different parenthesizations can lead to same values. For example, all the ways to group the addition with 5+1 in different positions give same value (144). Similarly, the other groupings give different values.

So, the four distinct values are 121, 122, 126, 144.

Therefore, the answer is 4. But let me check again if there's another value.

Wait, what if we do ( (2 * 3) + 1 ) * 4 * 5? This would be (6 +1)*20=7*20=140. Is this allowed? The numbers are in order 2, 3, 4, 5, 1. The addition here is between 2 and 3, but they are not adjacent in the original expression. Wait, original expression is 2 * 3 * 4 * 5 + 1. To do 2*3 +1, you have to group (2*3 +1). However, the +1 is at the end, so to have 2*3 +1, you need to have the addition between 3 and 1, but they are not adjacent. Hence, this is not allowed. Therefore, this is invalid.

Similarly, if you try to do 2*(3+1) *4*5, you need to have the +1 between 3 and 1, which are not adjacent. So no.

Therefore, 140 is invalid.

Alternatively, is there any other way?

What if we do ((2*3*4) *5) +1, which is same as original.

Or ((2 * (3 * 4)) * 5) +1, same as original.

Alternatively, ((2 * 3) * (4 * 5)) +1, same as original.

So no new values.

Therefore, yes, only four values: 121, 122, 126, 144.

Wait, but let me check if 122 is indeed a valid value.

Yes: 2 * (3 * (4 * 5) +1). Let me verify that.

Original expression: 2 * 3 * 4 * 5 +1. To compute 2 * (3 * (4 * 5) +1), you first compute 4*5=20, then add 1: 21, then multiply by 3: 63, then multiply by 2: 126. Wait, that gives 126, not 122. Wait, I think I made a mistake earlier.

Wait, earlier I thought that 2 * ( (3 * 4 * 5) +1 ) gives 122. Let's compute this:

First, compute 3 * 4 * 5 = 60, then add 1: 61, then multiply by 2: 122. Yes, that's correct. So that is 122.

But if I do 2 * (3 * (4 * 5 +1)), that's 4*5 +1 =21, 3*21=63, 2*63=126.

So there are two different parenthesizations that give different results: 122 and 126.

So, the four values are:

- 121 (original)

- 122 (2 * (3 * 4 * 5 +1))

- 126 (2 * (3 * (4 * 5 +1)))

- 144 ((2 * 3 * 4) * (5 + 1) etc.)

Therefore, four distinct values.

So the answer should be 4. However, I need to confirm if there are any other ways.

Let me think of another way: for example, (((2 * 3) + 1) * 4) * 5. But this would be adding 2 and 3, but they are not adjacent. Invalid.

Another invalid attempt: ((2 + 3) * 4 * 5) +1. Invalid because + is not between 2 and 3.

Thus, no.

Therefore, I think the answer is 4.

However, I recall that in some similar problems, the number might be higher. Let me think again.

Let me think of all possible parenthesizations:

The expression is 2 * 3 * 4 * 5 + 1. Let's list all the possible parenthesizations, considering the order of operations and where the addition can be.

First, the operations are:

Positions:

1. Between 2 and 3: M1

2. Between 3 and 4: M2

3. Between 4 and 5: M3

4. Between 5 and 1: A

So, total four operations.

The different ways to parenthesize this expression can be represented as different orders of applying operations.

The possible ways are determined by the different ways of choosing which operation to apply last.

For example:

1. Apply addition last: perform all multiplications first, then add 1. Value: 121.

2. Apply addition before some multiplication.

Let's consider the different scenarios where the addition is applied before some multiplications:

Case A: Apply addition at the end (op4).

Case B: Apply addition after op1 (between 2 and 3). But to do that, the addition would have to be between 2 and the rest. But how?

Wait, the addition is only between 5 and 1. To apply addition earlier, it has to be combined with some other operations.

For example:

- Apply op4 (addition) between 5 and 1, then multiply with 4: 4*(5 +1) =24, then multiply with 2*3=6: 6*24=144.

- Apply op4 between 4 and 1: Wait, no, op4 is between 5 and 1. To apply addition between 4 and 5, it would require changing op3 from multiplication to addition, which is not allowed.

Hence, the only way to apply the addition earlier is to apply it at op4 (between 5 and 1), then multiply with the rest.

Therefore, the addition can only be applied at op4, but the rest can be grouped in different ways.

Thus, the different groupings are determined by how the multiplications are grouped before applying the addition.

So, the different groupings are:

- Group all multiplications first: (2*3*4*5)+1=121

- Group the first three numbers, then multiply by (5+1): (2*3*4)*(5+1)=144

- Group the first two numbers, then multiply by (4*(5+1)): (2*3)*(4*(5+1))=6*24=144

- Group the first number, then multiply by (3*4*(5+1)): 2*(3*4*(5+1))=144

- Group the first and second numbers, then multiply by (4*5), then add 1: Wait, but that's the original.

Wait, but there's also the case where the addition is applied after some multiplications, like:

- (2*3)*(4*5) +1 = original

- 2*(3*4)*5 +1 = original

- 2*3*(4*5) +1 = original

So these don't give new values.

However, there's also:

- 2*(3*(4*5 +1)) which is 2*(3*(20+1))= 2*63=126

- 2*( (3*4)*5 +1 ) which is 2*(12*5 +1)=2*(60 +1)=122

So these are two different values.

Therefore, adding those two, we have:

121, 122, 126, 144.

Thus, 4 values.

Therefore, the answer should be 4. But let me check if there's another value from different grouping.

For example, what about ((2 * 3) * (4 + 5)) + 1. But as discussed earlier, this would require changing op3 from multiplication to addition, which is not allowed.

Alternatively, ((2 * (3 * (4 * 5))) +1 ) = (2 * 60) +1=121. Same as original.

So, no new value.

Thus, the total is four distinct values: 121, 122, 126, 144.

Therefore, the answer is 4.

But let me verify by considering that the problem says "inserting parentheses" and that the example gives 144. The question is from a competition, and sometimes these problems have more answers, so I want to be sure.

Wait, let's think about all possible values:

1. 121: original.

2. 122: 2*(3*(4*5) + 1)=122.

3. 126: 2*(3*(4*5 +1))=126.

4. 144: 2*(3*4*(5+1))=144, or (2*3*4)*(5+1)=144, etc.

Is there another value? Let me think.

If we do ((2 * 3) * 4 * 5) + 1 = 121.

Or, 2 * (3 * 4 * 5) +1 = 121? Wait, no, that would be 2*60 +1=121. Wait, but this is same as original? No, the original is 2*3*4*5 +1= 120 +1=121. If I do 2*(3*4*5) +1=2*60 +1=121, same value.

Wait, but if I do 2*(3*4*5) +1, is that a different parenthesization? Or is it same as the original?

In terms of grouping, it's 2*(3*4*5) +1, which is different from ((2*3)*4)*5 +1, but since multiplication is associative, it evaluates to the same value. So, even though the grouping is different, the value is same. Therefore, it doesn't contribute to new value.

Thus, the only additional values are 122, 126, and 144. Hence, total four values.

Therefore, I think the answer is 4.

However, to make sure, let me think of another approach. Since there are four operations, and the addition is one of them, the number of distinct values depends on when the addition is applied.

The addition can be applied either after all multiplications (value 121), or after some multiplications.

When applied after some multiplications, the number of possible ways is equal to the number of ways to split the multiplicative sequence into two parts, one of which is multiplied with the result of (5 +1).

As I did before, there are three such splits (excluding the full multiplication), giving values 122, 126, 144.

Hence, adding the original, total four values.

Yes, that makes sense.

Therefore, I conclude that the number of distinct values is 4.

**Final Answer**
\boxed{4}

To solve the problem, we are given the expression:

$$
2 \cdot 3 \cdot 4 \cdot 5 + 1
$$

We are to determine how many **distinct values** this expression can produce when **parentheses are inserted**, **without rearranging the order of the operands** (i.e., the numbers must remain in the order 2, 3, 4, 5, 1).

---

### Step 1: Understand the Original Expression

The original expression is interpreted as:

$$
(2 \cdot 3) \cdot (4 \cdot 5) + 1 = 120 + 1 = 121
$$

This is the default value when all multiplications are performed before the addition.

We are asked to insert **parentheses** in the expression to alter the **order of operations**, and to count how many **distinct values** can be obtained.

---

### Step 2: Consider the Structure

The expression contains four numbers and four **operations**: three multiplications and **one addition**. The key rule is that **you can only insert parentheses**, so you cannot rearrange the numbers or add new operations.

The addition is **only between 5 and 1**, as the original expression is:

$$
2 \cdot 3 \cdot 4 \cdot 5 + 1
$$

Therefore, **the only legal place for the addition** is between the product of the first four numbers and the final term, 1. But we are allowed to **insert parentheses** to **group parts of the multiplication** with the addition, effectively changing when the addition is performed.

---

### Step 3: Determine All Valid Parentheses Configurations

We will consider different ways to **group** the expression, focusing on where the addition occurs.

Some of the valid expressions (with parentheses) are:

1. **Original (default):**
   $$
   (((2 \cdot 3) \cdot 4) \cdot 5) + 1 = 121
   $$

2. **Group 5 and 1 before multiplying the rest:**
   $$
   (2 \cdot 3 \cdot 4) \cdot (5 + 1) = 24 \cdot 6 = 144
   $$

3. **Group 4 and 5 with 1:**
   $$
   (2 \cdot 3) \cdot (4 \cdot (5 + 1)) = 6 \cdot 24 = 144
   $$

4. **Group 3, 4, and 5 with 1:**
   $$
   2 \cdot (3 \cdot (4 \cdot 5 + 1)) = 2 \cdot (3 \cdot 21) = 2 \cdot 63 = 126
   $$

5. **Group the first three numbers with the result of the addition:**
   $$
   2 \cdot (3 \cdot (4 \cdot 5) + 1) = 2 \cdot (60 + 1) = 2 \cdot 61 = 122
   $$

---

### Step 4: Gather All Distinct Values

From the above configurations, we have computed the following results:

- **121** (original: all multiplications, then addition)
- **122** (2 * (3 * (4 * 5) + 1))
- **126** (2 * (3 * (4 * 5 + 1)))
- **144** (multiple distinct parenthesizations, all yield 144)

Each of these values is **unique** and comes from **different valid insertions of parentheses**, without rearranging operands or adding new operations.

---

### ✅ Final Conclusion

There are **4** distinct values that can be obtained by inserting parentheses in the expression **2·3·4·5 + 1**, considering only valid parenthesizations that preserve the order of operands and apply operations as specified.

$$
\boxed{4}
$$

Solution:
By the associative property of multiplication, it doesn't help to insert parentheses that specify the order of multiplication. For example, the associative property tells us that $(2\cdot(3\cdot 4))\cdot (5+1)$ is the same as $2\cdot3\cdot4\cdot (5+1)$. So the only way to get different values is to group +1 with a different number of factors. We get \begin{align*}
2\cdot 3 \cdot 4 \cdot (5 + 1) &= 144, \\
2\cdot 3 \cdot (4 \cdot 5 + 1) &= 126,\\
2\cdot (3 \cdot 4 \cdot 5 + 1) &= 122, \\
(2\cdot 3 \cdot 4 \cdot 5) + 1 \hphantom{)} &= 121.
\end{align*}In total there are $\boxed{4}$ possible values for the expression.