Explanation:
Okay, so there's this problem about six cars pulling up to a red light, one at a time. The light has three lanes: left-turn, straight-going, and right-turn. The question is asking how many ways the cars can stack up so that all three lanes are occupied. The cars are distinguishable, meaning each one is different, and they come in a fixed order. So the first car is different from the second, etc.

First, let me try to understand the problem. Each car has to choose one of the three lanes when it comes to the light. However, the problem says that all three lanes must be occupied. That means, after all six cars have pulled up, each of the three lanes has at least one car in it. So we need to count the number of ways the cars can be assigned to lanes such that none of the lanes are empty.

But also, since the cars are distinguishable and they arrive one at a time, the order in which they arrive matters? Wait, but does the order in which they arrive affect how they stack up? Let me check the note.

The note says that if the first car turns left and the second goes straight, that's different from the first car going straight and the second turning left. So yes, the order of the cars matters. Therefore, each car's choice of lane is a sequence, and the assignments are ordered. However, we need to count all such assignments where all three lanes are occupied.

So maybe this is similar to counting the number of functions from the set of cars to the set of lanes, with the constraint that each lane is used at least once. However, since the cars are distinguishable and the order matters, perhaps we need to think in terms of permutations or something else? Wait, no. Wait, when assigning each car to a lane, each car independently chooses a lane, but we have the constraint that all three lanes are used.

Wait, but since the cars are arriving one after another, perhaps each car's lane choice is a decision that can be made independently. So the total number of ways without any restrictions would be 3^6, since each car has three choices. But we need to subtract those cases where one or more lanes are empty.

But hold on, the problem mentions that the cars stack up. So maybe the order in which they arrive also affects the stacking? For example, if two cars go into the same lane, their order would matter in the stacking? Hmm. Wait, the problem says that the cars are distinguishable and pull up in a fixed order. So the order of the cars is fixed. So the first car is first, then the second, etc. But each car is assigned to a lane. However, the stacking up refers to the arrangement of cars in the lanes.

Wait, maybe the problem is that when the cars arrive, they enter the lane they are going to. So each lane has a sequence of cars. For example, the left-turn lane has some number of cars, each one arriving one after another. Similarly for the straight and right lanes. However, since the cars arrive in order, the order in the lanes is determined by the order they arrived.

Therefore, actually, each car is assigned to a lane, and the assignment is a function from cars to lanes, but with the constraint that all three lanes have at least one car. However, since the cars are distinguishable and the order is fixed, the assignment is a sequence of lane choices.

But the problem says "how many ways can the cars stack up so that all three lanes are occupied." So maybe it's equivalent to counting the number of functions from the six cars to the three lanes, with each lane having at least one car.

Therefore, this is similar to the inclusion-exclusion principle. The total number of functions is 3^6. Then subtract the ones that miss at least one lane.

So using inclusion-exclusion, the number of onto functions from 6 cars to 3 lanes is:

Total = 3^6 - C(3,1)*2^6 + C(3,2)*1^6 - C(3,3)*0^6

But since 0^6 is zero, the formula simplifies to:

3^6 - 3*2^6 + 3*1^6.

Calculating that:

3^6 is 729

2^6 is 64, so 3*64 is 192

1^6 is 1, so 3*1 is 3

So total is 729 - 192 + 3 = 729 - 192 is 537, plus 3 is 540.

So is the answer 540? But let me check again.

Wait, but maybe there's more to it? Because the problem mentions that the cars stack up. Maybe each lane's order matters? Wait, but if the cars are arriving in a fixed order, and each car is assigned to a lane, then the order of cars in each lane is determined by their arrival order. So for example, if two cars go into the left-turn lane, the first one that arrived is first in the lane, then the second.

But since the problem counts different orderings of cars as different, even if they are in the same lane, but since the cars are distinguishable and the arrival order is fixed, the assignment to lanes is just which lane each car goes to. So the total number of assignments where all three lanes are occupied is indeed the number of onto functions from 6 cars to 3 lanes, which is 540.

Wait, but maybe the problem wants to consider the order of cars in each lane? Wait, but the problem says that the cars are distinguishable and pull up in a fixed order. So the order of arrival is fixed. So if two cars go into the same lane, their order in the lane is determined by their arrival order. Therefore, different assignments would correspond to different ways of stacking up.

But if we think of the lanes as queues, where cars are added to the end of the queue. So for example, if car 1 goes to lane A and car 2 goes to lane A, then lane A has car 1 followed by car 2.

Therefore, the problem is equivalent to assigning each car to one of the three lanes, with the constraint that all lanes have at least one car. Since the order of arrival is fixed, each assignment corresponds to a unique stacking. Therefore, the answer is indeed the number of onto functions, which is 540.

Wait, but let me check again with another approach.

Suppose we think of the problem as first assigning the cars to the lanes with all lanes occupied, and then considering the order. But since the order is already determined by the arrival order, maybe it's just the same as assigning.

Alternatively, suppose we think of it as permutations with some constraints. But no, since each car is independently assigned to a lane.

Alternatively, maybe there's a misunderstanding here. Maybe the problem is not about assigning each car to a lane, but about arranging them in the lanes such that each lane has at least one car. But since they arrive one by one, the order of the cars is fixed.

Wait, but perhaps the problem is that the cars are stacking up in the lanes, so each lane is a line of cars, and the order of cars in each lane is the order they arrived. Therefore, the entire configuration is determined by which cars are in which lane.

Therefore, the problem reduces to counting the number of ways to partition six distinguishable cars into three non-empty subsets (for the three lanes), multiplied by the number of ways to assign these subsets to the lanes. Wait, but since the lanes are distinct (left-turn, straight, right-turn), we don't need to multiply by anything else.

So, the number is equal to the number of onto functions from the six cars to the three lanes, which is 3! * S(6,3), where S(6,3) is the Stirling numbers of the second kind.

But let me compute that. The formula for onto functions is indeed 3! * S(6,3).

Let me recall that S(n,k) is the number of ways to partition a set of n objects into k non-empty subsets. Then, multiplying by k! gives the number of onto functions, since we assign each subset to a specific lane.

Alternatively, the formula using inclusion-exclusion is as above: 3^6 - 3*2^6 + 3*1^6.

Let me compute S(6,3).

The value of S(6,3) can be calculated using the recurrence relation:

S(n, k) = S(n-1, k-1) + k*S(n-1, k)

But maybe I can recall that S(6,3) is 90. Then 3! * 90 = 6 * 90 = 540, which matches the previous number.

Alternatively, if I don't recall the value, I can compute it.

Alternatively, the inclusion-exclusion formula gives 540, so that's consistent.

So then, is the answer 540?

But let me think again if there's something missing here.

Wait, the problem says "stack up so that all three lanes are occupied". So maybe the order of cars in each lane matters? But since the cars arrive in a fixed order, the order in the lane is determined by their arrival. So the assignment of cars to lanes is all that's needed. Therefore, different assignments correspond to different stackings.

Alternatively, maybe the problem is considering the order of cars in each lane as part of the stacking. But if the cars are distinguishable and arrive one at a time, then the order in the lane is fixed by the arrival order. For example, if two cars go into the left-turn lane, the first one that arrived will be ahead of the second one in the lane.

Therefore, if we just assign each car to a lane, the entire configuration is determined. Therefore, the total number of ways is the number of onto functions, which is 540.

But wait, let me check with a smaller case. Let's say there are 3 cars and 3 lanes. How many ways?

Each car must go to a different lane. Since the cars are distinguishable, it's 3! = 6. According to inclusion-exclusion, 3^3 - 3*2^3 + 3*1^3 = 27 - 24 + 3 = 6. Correct.

Another test case: 4 cars, 3 lanes. The number of onto functions is 3^4 - 3*2^4 + 3*1^4 = 81 - 48 + 3 = 36.

Alternatively, using Stirling numbers: S(4,3) = 6, so 3! * 6 = 36. Correct.

So our formula works. Therefore, for 6 cars, 540 is the correct answer.

Wait, but the problem says "stack up". Maybe there's something else? For example, when cars are in the same lane, they stack up in some order? But since the cars arrive one at a time, the stacking order is already determined. Therefore, there's no further permutations needed.

Alternatively, maybe the problem is considering that cars in the same lane are arranged in some order? But no, the problem says that the cars pull up in a fixed order. So the first car is first, then the second, etc. So the order of the cars in each lane is the same as their arrival order.

Therefore, I think the answer is indeed 540.

But wait, let me check once more.

Suppose there are 2 cars and 2 lanes. The number of ways to have both lanes occupied is 2! = 2. According to inclusion-exclusion: 2^2 - 2*1^2 = 4 - 2 = 2. Correct.

Another way: For each car, assign to a lane, subtract those assignments where one lane is empty.

Therefore, returning, with 6 cars and 3 lanes, the answer is 540. Therefore, I think that is the answer.

However, just to be thorough, let me think if the problem has any other constraints.

The problem says "stack up so that all three lanes are occupied". It doesn't specify any other constraints, like cars can't be in certain positions or anything. The only thing is that each lane has at least one car.

Therefore, the answer is 540. So I think \boxed{540} is the answer.

But wait, hold on. Let me check once more.

Alternatively, maybe the problem is about arranging the cars in the lanes considering the order in the lane? Wait, but the problem says "how many ways can the cars stack up". If "stack up" implies that the order of cars in each lane matters, then perhaps we need to consider permutations?

But in that case, maybe the problem is equivalent to: Assign each car to a lane (with all lanes non-empty), and then for each lane, arrange the cars in that lane in some order. But since the cars arrive in a fixed order, the order in the lane is determined by arrival.

Wait, but maybe the problem is considering that the cars can be arranged in the lane in any order, not necessarily the arrival order? But that seems inconsistent with the note given.

The note says that if the first car turns left and the second goes straight, this is different from the first car going straight and the second turning left. So that indicates that the order of the cars is important, but only in terms of the sequence of their choices.

So the assignment is a sequence of lane choices for each car. So if you assign car 1 to lane A, car 2 to lane B, etc., that's different from assigning car 1 to B, car 2 to A. So the order of the cars matters because they are distinguishable and arrive in a fixed order.

Therefore, the problem is equivalent to counting the number of sequences of length 6, where each element is one of three values (left, straight, right), such that all three values appear at least once.

Which is exactly the number of onto functions from 6 cars to 3 lanes, which is 540.

So I think that is the answer. Therefore, the answer is 540.

I think I have considered multiple angles and tested with smaller numbers. So I think it's safe to go with 540.

**Final Answer**
\boxed{540}

To solve the problem, we must determine how many ways six distinguishable cars can "stack up" at a red light with three lanes—left-turn, straight, and right-turn—such that *all three lanes are occupied*. Here's how we approach it step by step:

---

### 1. **Understanding the Problem**

- **6 cars** arrive one at a time in a **fixed order**, meaning the order in which the cars are assigned to lanes matters.
- **Each car is distinguishable**, meaning swapping one car with another results in a different configuration.
- The **light has three lanes**: left-turn, straight, and right-turn.
- The goal is to count all such **assignments of cars to lanes** such that **each lane has at least one car**.

This is a classic **onto function** problem where we are assigning cars (elements of a set) to lanes (elements of a set of 3), with the constraint that **each lane gets at least one car**.

---

### 2. **Total Assignments Without Restriction**

If each of the 6 cars independently chooses one of the three lanes:

$$
\text{Total assignments} = 3^6 = 729
$$

This counts all possible assignments of cars to lanes, with no restriction that each lane must have at least one car.

---

### 3. **Subtract Assignments Where One or More Lanes Are Empty**

To enforce that **all lanes are occupied**, we use the **principle of inclusion-exclusion**.

- **Step 1:** Subtract assignments where **at least one lane is empty**.

We consider the number of ways that one particular lane is left empty (i.e., all 6 cars are assigned to the remaining two lanes). The number of such assignments is:

$$
\text{Ways to leave one lane empty} = 2^6 = 64
$$

There are $ \binom{3}{1} = 3 $ ways to choose which lane is left empty, so:

$$
3 \times 2^6 = 3 \times 64 = 192
$$

- **Step 2:** Add back assignments where **two lanes are empty**, since they were subtracted twice.

If two lanes are empty, all 6 cars must go into the remaining one lane. There is only $ 1^6 = 1 $ way to do this, and there are $ \binom{3}{2} = 3 $ ways to choose which two lanes are empty.

$$
3 \times 1^6 = 3 \times 1 = 3
$$

- **Step 3:** Finally, subtract the assignment where **all three lanes are empty**, but this is not possible since all cars must go somewhere.

So, using inclusion-exclusion:

$$
\text{Valid assignments} = 3^6 - 3 \times 2^6 + 3 \times 1^6 = 729 - 192 + 3 = 540
$$

---

### 4. **Alternative Perspective (Stirling Numbers)**

Another way to think about this is using **Stirling numbers of the second kind**, $ S(n, k) $, which count the number of ways to partition $ n $ elements into $ k $ non-empty subsets. Since the lanes are **distinct**, we multiply by $ k! $ to assign the subsets to the lanes.

$$
S(6, 3) = 90 \quad \Rightarrow \quad 3! \times 90 = 6 \times 90 = 540
$$

This confirms our earlier result.

---

### 5. **Conclusion**

Since the cars are distinguishable and arrive in a fixed order, and the lanes are **distinct**, every valid assignment corresponds to a unique way the cars can stack up (in the sense of being grouped into lanes with each lane having at least one car). Hence, the number of such arrangements is:

$$
\boxed{540}
$$

[Assistant]
Explanation:
Okay, so there's this problem about six cars pulling up to a red light, one at a time. The light has three lanes: left-turn, straight-going, and right-turn. The question is asking how many ways the cars can stack up so that all three lanes are occupied. The cars are distinguishable, meaning each one is different, and they come in a fixed order. So the first car is different from the second, etc.

First, let me try to understand the problem. Each car has to choose one of the three lanes when it comes to the light. However, the problem says that all three lanes must be occupied. That means, after all six cars have pulled up, each of the three lanes has at least one car in it. So we need to count the number of ways the cars can be assigned to lanes such that none of the lanes are empty.

But also, since the cars are distinguishable and they arrive one at a time, the order in which they arrive matters? Wait, but does the order in which they arrive affect how they stack up? Let me check the note.

The note says that if the first car turns left and the second goes straight, that's different from the first car going straight and the second turning left. So yes, the order of the cars matters. Therefore, each car's choice of lane is a sequence, and the assignments are ordered. However, we need to count all such assignments where all three lanes are occupied.

So maybe this is similar to counting the number of functions from the set of cars to the set of lanes, with the constraint that each lane is used at least once. However, since the cars are distinguishable and the order matters, perhaps we need to think in terms of permutations or something else? Wait, no. Wait, when assigning each car to a lane, each car independently chooses a lane, but we have the constraint that all three lanes are used.

Wait, but since the cars are arriving one after another, perhaps each car's lane choice is a decision that can be made independently. So the total number of ways without any restrictions would be 3^6, since each car has three choices. But we need to subtract those cases where one or more lanes are empty.

But hold on, the problem mentions that the cars stack up. So maybe the order in which they arrive also affects the stacking? For example, if two cars go into the same lane, their order would matter in the stacking? Hmm. Wait, the problem says that the cars are distinguishable and pull up in a fixed order. So the order of the cars is fixed. So the first car is first, then the second, etc. But each car is assigned to a lane. However, the stacking up refers to the arrangement of cars in the lanes.

Wait, maybe the problem is that when the cars arrive, they enter the lane they are going to. So each lane has a sequence of cars. For example, the left-turn lane has some number of cars, each one arriving one after another. Similarly for the straight and right lanes. However, since the cars arrive in order, the order in the lanes is determined by the order they arrived.

Therefore, actually, each car is assigned to a lane, and the assignment is a function from cars to lanes, but with the constraint that all three lanes have at least one car. However, since the cars are distinguishable and the order is fixed, the assignment is a sequence of lane choices.

But the problem says "how many ways can the cars stack up so that all three lanes are occupied." So maybe it's equivalent to counting the number of functions from the six cars to the three lanes, with each lane having at least one car.

Therefore, this is similar to the inclusion-exclusion principle. The total number of functions is 3^6. Then subtract the ones that miss at least one lane.

So using inclusion-exclusion, the number of onto functions from 6 cars to 3 lanes is:

Total = 3^6 - C(3,1)*2^6 + C(3,2)*1^6 - C(3,3)*0^6

But since 0^6 is zero, the formula simplifies to:

3^6 - 3*2^6 + 3*1^6.

Calculating that:

3^6 is 729

2^6 is 64, so 3*64 is 192

1^6 is 1, so 3*1 is 3

So total is 729 - 192 + 3 = 729 - 192 is 537, plus 3 is 540.

So is the answer 540? But let me check again.

Wait, but maybe there's more to it? Because the problem mentions that the cars stack up. Maybe each lane's order matters? Wait, but if the cars are arriving in a fixed order, and each car is assigned to a lane, then the order of cars in each lane is determined by their arrival order. So for example, if two cars go into the left-turn lane, the first one that arrived is first in the lane, then the second.

But since the problem counts different orderings of cars as different, even if they are in the same lane, but since the cars are distinguishable and the arrival order is fixed, the assignment to lanes is just which lane each car goes to. So the total number of assignments where all three lanes are occupied is indeed the number of onto functions from 6 cars to 3 lanes, which is 540.

Wait, but maybe the problem wants to consider the order of cars in each lane? Wait, but the problem says that the cars are distinguishable and pull up in a fixed order. So the order of arrival is fixed. So if two cars go into the same lane, their order in the lane is determined by their arrival order. Therefore, different assignments would correspond to different ways of stacking up.

But if we think of the lanes as queues, where cars are added to the end of the queue. So for example, if car 1 goes to lane A and car 2 goes to lane A, then lane A has car 1 followed by car 2.

Therefore, the problem is equivalent to assigning each car to one of the three lanes, with the constraint that all lanes have at least one car. Since the order of arrival is fixed, each assignment corresponds to a unique stacking. Therefore, the answer is indeed the number of onto functions, which is 540.

Wait, but let me check again with another approach.

Suppose we think of the problem as first assigning the cars to the lanes with all lanes occupied, and then considering the order. But since the order is already determined by the arrival order, maybe it's just the same as assigning.

Alternatively, suppose we think of it as permutations with some constraints. But no, since each car is independently assigned to a lane.

Alternatively, maybe there's a misunderstanding here. Maybe the problem is not about assigning each car to a lane, but about arranging them in the lanes such that each lane has at least one car. But since they arrive one by one, the order of the cars is fixed.

Wait, but perhaps the problem is that the cars are stacking up in the lanes, so each lane is a line of cars, and the order of cars in each lane is the order they arrived. Therefore, the entire configuration is determined by which cars are in which lane.

Therefore, the problem reduces to counting the number of ways to partition six distinguishable cars into three non-empty subsets (for the three lanes), multiplied by the number of ways to assign these subsets to the lanes. Wait, but since the lanes are distinct (left-turn, straight, right-turn), we don't need to multiply by anything else.

So, the number is equal to the number of onto functions from the six cars to the three lanes, which is 3! * S(6,3), where S(6,3) is the Stirling numbers of the second kind.

But let me compute that. The formula for onto functions is indeed 3! * S(6,3).

Let me recall that S(n,k) is the number of ways to partition a set of n objects into k non-empty subsets. Then, multiplying by k! gives the number of onto functions, since we assign each subset to a specific lane.

Alternatively, the formula using inclusion-exclusion is as above: 3^6 - 3*2^6 + 3*1^6.

Let me compute S(6,3).

The value of S(6,3) can be calculated using the recurrence relation:

S(n, k) = S(n-1, k-1) + k*S(n-1, k)

But maybe I can recall that S(6,3) is 90. Then 3! * 90 = 6 * 90 = 540, which matches the previous number.

Alternatively, if I don't recall the value, I can compute it.

Alternatively, the inclusion-exclusion formula gives 540, so that's consistent.

So then, is the answer 540?

But let me think again if there's something missing here.

Wait, the problem says "stack up so that all three lanes are occupied". So maybe the order of cars in each lane matters? But since the cars arrive in a fixed order, the order in the lane is determined by their arrival. So the assignment of cars to lanes is all that's needed. Therefore, different assignments correspond to different stackings.

Alternatively, maybe the problem is considering the order of cars in each lane as part of the stacking. But if the cars are distinguishable and arrive one at a time, then the order in the lane is fixed by the arrival order. For example, if two cars go into the left-turn lane, the first one that arrived will be ahead of the second one in the lane.

Therefore, if we just assign each car to a lane, the entire configuration is determined. Therefore, the total number of ways is the number of onto functions, which is 540.

But wait, let me check with a smaller case. Let's say there are 3 cars and 3 lanes. How many ways?

Each car must go to a different lane. Since the cars are distinguishable, it's 3! = 6. According to inclusion-exclusion, 3^3 - 3*2^3 + 3*1^3 = 27 - 24 + 3 = 6. Correct.

Another test case: 4 cars, 3 lanes. The number of onto functions is 3^4 - 3*2^4 + 3*1^4 = 81 - 48 + 3 = 36.

Alternatively, using Stirling numbers: S(4,3) = 6, so 3! * 6 = 36. Correct.

So our formula works. Therefore, for 6 cars, 540 is the correct answer.

Wait, but the problem says "stack up". Maybe there's something else? For example, when cars are in the same lane, they stack up in some order? But since the cars arrive one at a time, the stacking order is already determined. Therefore, there's no further permutations needed.

Alternatively, maybe the problem is considering that cars in the same lane are arranged in some order? But no, the problem says that the cars pull up in a fixed order. So the first car is first, then the second, etc. So the order of the cars in each lane is the same as their arrival order.

Therefore, I think the answer is indeed 540.

But wait, let me check once more.

Suppose there are 2 cars and 2 lanes. The number of ways to have both lanes occupied is 2! = 2. According to inclusion-exclusion: 2^2 - 2*1^2 = 4 - 2 = 2. Correct.

Another way: For each car, assign to a lane, subtract those assignments where one lane is empty.

Therefore, returning, with 6 cars and 3 lanes, the answer is 540. Therefore, I think that is the answer.

However, just to be thorough, let me think if the problem has any other constraints.

The problem says "stack up so that all three lanes are occupied". It doesn't specify any other constraints, like cars can't be in certain positions or anything. The only thing is that each lane has at least one car.

Therefore, the answer is 540. So I think \boxed{540} is the answer.

But wait, hold on. Let me check once more.

Alternatively, maybe the problem is about arranging the cars in the lanes considering the order in the lane? Wait, but the problem says "how many ways can the cars stack up". If "stack up" implies that the order of cars in each lane matters, then perhaps we need to consider permutations?

But in that case, maybe the problem is equivalent to: Assign each car to a lane (with all lanes non-empty), and then for each lane, arrange the cars in that lane in some order. But since the cars arrive in a fixed order, the order in the lane is determined by arrival.

Wait, but maybe the problem is considering that the cars can be arranged in the lane in any order, not necessarily the arrival order? But that seems inconsistent with the note given.

The note says that if the first car turns left and the second goes straight, this is different from the first car going straight and the second turning left. So that indicates that the order of the cars is important, but only in terms of the sequence of their choices.

So the assignment is a sequence of lane choices for each car. So if you assign car 1 to lane A, car 2 to lane B, etc., that's different from assigning car 1 to B, car 2 to A. So the order of the cars matters because they are distinguishable and arrive in a fixed order.

Therefore, the problem is equivalent to counting the number of sequences of length 6, where each element is one of three values (left, straight, right), such that all three values appear at least once.

Which is exactly the number of onto functions from 6 cars to 3 lanes, which is 540.

So I think that is the answer. Therefore, the answer is 540.

I think I have considered multiple angles and tested with smaller numbers. So I think it's safe to go with 540.

**Final Answer**
\boxed{540}

To solve the problem, we must determine how many ways six distinguishable cars can "stack up" at a red light with three lanes—left-turn, straight, and right-turn—such that *all three lanes are occupied*. Here's how we approach it step by step:

---

### 1. **Understanding the Problem**

- **6 cars** arrive one at a time in a **fixed order**, meaning the order in which the cars are assigned to lanes matters.
- **Each car is distinguishable**, meaning swapping one car with another results in a different configuration.
- The **light has three lanes**: left-turn, straight, and right-turn.
- The goal is to count all such **assignments of cars to lanes** such that **each lane has at least one car**.

This is a classic **onto function** problem where we are assigning cars (elements of a set) to lanes (elements of a set of 3), with the constraint that **each lane gets at least one car**.

---

### 2. **Total Assignments Without Restriction**

If each of the 6 cars independently chooses one of the three lanes:

$$
\text{Total assignments} = 3^6 = 729
$$

This counts all possible assignments of cars to lanes, with no restriction that each lane must have at least one car.

---

### 3. **Subtract Assignments Where One or More Lanes Are Empty**

To enforce that **all lanes are occupied**, we use the **principle of inclusion-exclusion**.

- **Step 1:** Subtract assignments where **at least one lane is empty**.

We consider the number of ways that one particular lane is left empty (i.e., all 6 cars are assigned to the remaining two lanes). The number of such assignments is:

$$
\text{Ways to leave one lane empty} = 2^6 = 64
$$

There are $ \binom{3}{1} = 3 $ ways to choose which lane is left empty, so:

$$
3 \times 2^6 = 3 \times 64 = 192
$$

- **Step 2:** Add back assignments where **two lanes are empty**, since they were subtracted twice.

If two lanes are empty, all 6 cars must go into the remaining one lane. There is only $ 1^6 = 1 $ way to do this, and there are $ \binom{3}{2} = 3 $ ways to choose which two lanes are empty.

$$
3 \times 1^6 = 3 \times 1 = 3
$$

- **Step 3:** Finally, subtract the assignment where **all three lanes are empty**, but this is not possible since all cars must go somewhere.

So, using inclusion-exclusion:

$$
\text{Valid assignments} = 3^6 - 3 \times 2^6 + 3 \times 1^6 = 729 - 192 + 3 = 540
$$

---

### 4. **Alternative Perspective (Stirling Numbers)**

Another way to think about this is using **Stirling numbers of the second kind**, $ S(n, k) $, which count the number of ways to partition $ n $ elements into $ k $ non-empty subsets. Since the lanes are **distinct**, we multiply by $ k! $ to assign the subsets to the lanes.

$$
S(6, 3) = 90 \quad \Rightarrow \quad 3! \times 90 = 6 \times 90 = 540
$$

This confirms our earlier result.

---

### 5. **Conclusion**

Since the cars are distinguishable and arrive in a fixed order, and the lanes are **distinct**, every valid assignment corresponds to a unique way the cars can stack up (in the sense of being grouped into lanes with each lane having at least one car). Hence, the number of such arrangements is:

$$
\boxed{540}
$$

Solution: We count the number of ways that some lane can be left empty, and subtract from the total number, $3^6=729$ because each driver has three choices. Suppose the left-turn lane is left empty. Then each driver is limited to 2 choices, and there are $2^6$ ways to leave the left-turn lane empty. The same logic gives $2^6$ ways to leave the center lane and right-turn lane open. But we have double-counted the situations where two lanes are left empty. Fortunately, since each driver must go into the third lane, there are only 3 situations we have over-counted. This leaves $3\cdot2^6-3=189$ ways to leave at least one lane unoccupied, or $729-189=\boxed{540}$ ways to occupy every lane.


Explanation:
Okay, so I need to find the maximum value of the function 4(x + 7)(2 - x) for all real numbers x. Let me think about how to approach this.

First, maybe I should expand the expression to make it easier to work with. Let me try that.

Starting with 4(x + 7)(2 - x). Let me first multiply out the two binomials (x + 7)(2 - x).

Multiplying (x + 7)(2 - x) can be done using the distributive property, also known as the FOIL method for binomials.

So, first terms: x * 2 = 2x
Outer terms: x * (-x) = -x²
Inner terms: 7 * 2 = 14
Last terms: 7 * (-x) = -7x

Now, adding all those up: 2x - x² + 14 - 7x.

Combine like terms: 2x - 7x is -5x, so we have -x² -5x +14.

Therefore, the original expression is 4*(-x² -5x +14). Let me distribute the 4 into the parentheses:

4*(-x²) + 4*(-5x) + 4*14 = -4x² -20x +56.

So, the function simplifies to f(x) = -4x² -20x +56.

Now, this is a quadratic function in standard form, which is f(x) = ax² + bx + c. Since the coefficient of x² is negative (-4), the parabola opens downward. That means the function has a maximum value at its vertex.

Therefore, the maximum value occurs at the vertex of this parabola. To find the x-coordinate of the vertex, I can use the formula x = -b/(2a).

In this case, a = -4 and b = -20. Plugging into the formula:

x = -(-20)/(2*(-4)) = 20 / (-8) = -2.5.

Wait, that gives me x = -2.5. Let me verify that again.

Yes, x = -b/(2a) = -(-20)/(2*(-4)) = 20 / (-8) = -2.5. Yes, that's correct.

So, the x-coordinate of the vertex is -2.5. To find the maximum value, I need to plug this back into the function f(x) to find the corresponding y-value.

Let me compute f(-2.5):

First, f(x) = -4x² -20x +56.

So, f(-2.5) = -4*(-2.5)² -20*(-2.5) +56.

Calculating each term step by step:

First term: (-2.5)² = 6.25, so -4*6.25 = -25.

Second term: -20*(-2.5) = 50.

Third term: 56.

Adding them all together: -25 + 50 + 56 = ( -25 + 50 ) + 56 = 25 + 56 = 81.

Wait, so the maximum value is 81? Let me double-check my calculations to make sure I didn't make an arithmetic error.

First, (-2.5)^2 is indeed 6.25. Multiply by -4: 6.25 * -4 = -25. Correct.

Then, -20 * (-2.5) is positive 50. Correct. Then adding 56. So total is -25 + 50 = 25, then 25 + 56 = 81. Yes, that seems correct.

Alternatively, maybe I can check using another method to confirm. Let me try completing the square.

Starting from f(x) = -4x² -20x +56.

First, factor out the coefficient of x² from the first two terms:

f(x) = -4(x² + 5x) + 56.

Now, to complete the square inside the parentheses. The coefficient of x is 5. Take half of 5, which is 2.5, and square it, which is 6.25.

So, add and subtract this inside the parentheses:

f(x) = -4[ (x² + 5x + 6.25 - 6.25) ] + 56

This becomes:

-4[ (x + 2.5)^2 - 6.25 ] + 56

Distribute the -4:

-4(x + 2.5)^2 + 25 + 56

Combine constants:

25 + 56 = 81

Therefore, f(x) = -4(x + 2.5)^2 + 81.

So, the maximum value is 81, occurring at x = -2.5. Therefore, that confirms my earlier calculation.

Alternatively, maybe I can use calculus. Taking the derivative of f(x) and setting it to zero.

But since this is a quadratic, calculus might be overkill, but let me try it for verification.

f(x) = -4x² -20x +56

f'(x) = -8x -20

Set derivative equal to zero:

-8x -20 = 0

-8x = 20

x = 20 / (-8) = -2.5

Same result. Then plug back into f(x), which gives 81 as before.

So, all methods lead to 81 as the maximum value.

Alternatively, maybe I can think of the original expression without expanding. The original function is 4(x + 7)(2 - x). Let me think if there's another way to find the maximum value.

Notice that (x + 7)(2 - x) can be rewritten as -(x + 7)(x - 2). Which is similar to a quadratic in terms of x. But since we already have that, maybe another approach is to think of it as a product of two linear terms and find its maximum.

Alternatively, use the vertex formula for a quadratic in factored form. Wait, but the standard vertex formula is for the standard form. However, if I can write the function in the form a(x - h)^2 + k, then k is the maximum or minimum.

But since we already did that earlier, it's the same as completing the square.

Alternatively, since it's a quadratic, the maximum occurs at the vertex, so x = -b/(2a) as we did earlier.

Alternatively, maybe using substitution. Let me think. Let me consider variables. Let me set y = x + 7 and z = 2 - x. Then, the product is 4yz. But I need to maximize yz. However, y and z are related because y = x + 7 and z = 2 - x. Therefore, adding them together: y + z = x + 7 + 2 - x = 9. So, y + z = 9.

Therefore, we need to maximize the product yz given that y + z = 9.

This is a classic optimization problem where for two numbers with a fixed sum, their product is maximized when they are equal. Wait, but that would be if they are positive. However, here y and z could be positive or negative? Let me check.

But since y = x + 7 and z = 2 - x, depending on the value of x, y and z can be positive or negative. However, if we want to maximize yz, given that y + z = 9.

But the maximum of yz when y + z is constant is when y = z = 4.5. However, is that possible? Let me check.

If y = z = 4.5, then y + z = 9, which is correct. Then, y = x + 7 = 4.5 => x = 4.5 - 7 = -2.5. Similarly, z = 2 - x = 4.5 => x = 2 - 4.5 = -2.5. So, that's consistent. Therefore, the maximum occurs at x = -2.5, which gives y = z = 4.5. Therefore, the maximum product yz is (4.5)^2 = 20.25. Then, multiplying by 4 gives 4 * 20.25 = 81. Which again gives 81.

So, this method also gives the same result. So that's a nice check.

Therefore, all methods agree that the maximum value is 81.

Alternatively, maybe I can use a graphing approach. If I imagine the graph of the function f(x) = 4(x + 7)(2 - x), which is a parabola opening downward, since the coefficient of x² is negative. The roots of the function are at x = -7 and x = 2. So, the parabola crosses the x-axis at -7 and 2, and opens downward. The vertex is halfway between the roots? Wait, no, the vertex is at the midpoint between the roots only if the parabola is symmetric. But since this is a standard quadratic, the axis of symmetry is halfway between the roots.

Wait, the roots are at x = -7 and x = 2. The midpoint between -7 and 2 is (-7 + 2)/2 = (-5)/2 = -2.5, which is the same x-coordinate as the vertex we found earlier. Therefore, the vertex is at x = -2.5, which is exactly halfway between the roots. Therefore, the maximum value is at x = -2.5, and substituting back gives 81, as we found.

Therefore, all methods confirm that the maximum value is 81.

But let me confirm once again by plugging in some values around x = -2.5 to see whether the function is indeed maximum there.

For example, let's take x = -2.5, which gives f(x) = 81.

Take x = -3:

f(-3) = 4(-3 + 7)(2 - (-3)) = 4(4)(5) = 4*20 = 80.

Which is less than 81.

Take x = -2:

f(-2) = 4(-2 + 7)(2 - (-2)) = 4(5)(4) = 4*20 = 80.

Also 80, less than 81.

Take x = -2.5:

f(-2.5) = 81.

Take x = -2.6:

f(-2.6) = 4(-2.6 + 7)(2 - (-2.6)) = 4(4.4)(4.6). Let me calculate that:

4 * 4.4 * 4.6. Let's compute 4.4 * 4.6 first.

Note that 4.4 * 4.6 = (4 + 0.4)(4 + 0.6) = 4*4 + 4*0.6 + 0.4*4 + 0.4*0.6 = 16 + 2.4 + 1.6 + 0.24 = 16 + 4 + 0.24 = 20.24.

Then multiply by 4: 20.24 * 4 = 80.96. Which is approximately 81, but slightly less.

Similarly, take x = -2.4:

f(-2.4) = 4(-2.4 + 7)(2 - (-2.4)) = 4(4.6)(4.4). Which is same as above, 4 * 4.4 * 4.6 = 80.96.

Therefore, confirming that at x = -2.5, the value is 81, which is indeed higher than the values at x = -2.6 and x = -2.4.

Therefore, this numerical check also supports that the maximum is 81.

Alternatively, if I take x = -2.5 + h and compute f(x) for small h, the function should have a maximum there.

But given that all methods lead to the same answer, and multiple checks confirm it, I can be confident that the maximum value is 81.

So, the answer is 81.

**Final Answer**
The maximum value is \boxed{81}.

To determine the maximum value of the expression
$$
4(x + 7)(2 - x)
$$
over all real numbers $ x $, we proceed by analyzing and simplifying the expression.

---

### Step 1: Expand the Product

We first simplify the expression:
$$
4(x + 7)(2 - x)
$$

Expanding the binomial:
$$
(x + 7)(2 - x) = -x^2 - 5x + 14
$$

Now multiply by 4:
$$
4(-x^2 - 5x + 14) = -4x^2 - 20x + 56
$$

So, the function becomes:
$$
f(x) = -4x^2 - 20x + 56
$$

This is a **quadratic function** with a **negative leading coefficient** ($-4$), which means the parabola opens **downward**. Therefore, the function has a **maximum value** at its **vertex**.

---

### Step 2: Find the Vertex

For a quadratic of the form $ f(x) = ax^2 + bx + c $, the x-coordinate of the vertex is given by:
$$
x = -\frac{b}{2a}
$$

Here, $ a = -4 $ and $ b = -20 $, so:
$$
x = -\frac{-20}{2(-4)} = \frac{20}{-8} = -2.5
$$

---

### Step 3: Evaluate the Function at the Vertex

To find the maximum value, evaluate $ f(-2.5) $:
$$
f(-2.5) = -4(-2.5)^2 - 20(-2.5) + 56
$$

Compute step-by-step:
- $ (-2.5)^2 = 6.25 $
- $ -4(6.25) = -25 $
- $ -20(-2.5) = 50 $
- So:
$$
f(-2.5) = -25 + 50 + 56 = 81
$$

---

### Step 4: Verify by Completing the Square (Optional)

We can verify this result by rewriting:
$$
f(x) = -4x^2 - 20x + 56
$$
as:
$$
f(x) = -4(x^2 + 5x) + 56
$$
Complete the square inside the parentheses:
$$
x^2 + 5x = (x + 2.5)^2 - 6.25
$$
So:
$$
f(x) = -4\left[(x + 2.5)^2 - 6.25\right] + 56 = -4(x + 2.5)^2 + 25 + 56 = -4(x + 2.5)^2 + 81
$$
This confirms the function achieves its **maximum value of 81** at $ x = -2.5 $, since the square term is always non-positive and minimized when it equals zero.

---

### Final Answer

The maximum value of the function is:

$$
\boxed{81}
$$

[Assistant]
Explanation:
Okay, so I need to find the maximum value of the function 4(x + 7)(2 - x) for all real numbers x. Let me think about how to approach this.

First, maybe I should expand the expression to make it easier to work with. Let me try that.

Starting with 4(x + 7)(2 - x). Let me first multiply out the two binomials (x + 7)(2 - x).

Multiplying (x + 7)(2 - x) can be done using the distributive property, also known as the FOIL method for binomials.

So, first terms: x * 2 = 2x
Outer terms: x * (-x) = -x²
Inner terms: 7 * 2 = 14
Last terms: 7 * (-x) = -7x

Now, adding all those up: 2x - x² + 14 - 7x.

Combine like terms: 2x - 7x is -5x, so we have -x² -5x +14.

Therefore, the original expression is 4*(-x² -5x +14). Let me distribute the 4 into the parentheses:

4*(-x²) + 4*(-5x) + 4*14 = -4x² -20x +56.

So, the function simplifies to f(x) = -4x² -20x +56.

Now, this is a quadratic function in standard form, which is f(x) = ax² + bx + c. Since the coefficient of x² is negative (-4), the parabola opens downward. That means the function has a maximum value at its vertex.

Therefore, the maximum value occurs at the vertex of this parabola. To find the x-coordinate of the vertex, I can use the formula x = -b/(2a).

In this case, a = -4 and b = -20. Plugging into the formula:

x = -(-20)/(2*(-4)) = 20 / (-8) = -2.5.

Wait, that gives me x = -2.5. Let me verify that again.

Yes, x = -b/(2a) = -(-20)/(2*(-4)) = 20 / (-8) = -2.5. Yes, that's correct.

So, the x-coordinate of the vertex is -2.5. To find the maximum value, I need to plug this back into the function f(x) to find the corresponding y-value.

Let me compute f(-2.5):

First, f(x) = -4x² -20x +56.

So, f(-2.5) = -4*(-2.5)² -20*(-2.5) +56.

Calculating each term step by step:

First term: (-2.5)² = 6.25, so -4*6.25 = -25.

Second term: -20*(-2.5) = 50.

Third term: 56.

Adding them all together: -25 + 50 + 56 = ( -25 + 50 ) + 56 = 25 + 56 = 81.

Wait, so the maximum value is 81? Let me double-check my calculations to make sure I didn't make an arithmetic error.

First, (-2.5)^2 is indeed 6.25. Multiply by -4: 6.25 * -4 = -25. Correct.

Then, -20 * (-2.5) is positive 50. Correct. Then adding 56. So total is -25 + 50 = 25, then 25 + 56 = 81. Yes, that seems correct.

Alternatively, maybe I can check using another method to confirm. Let me try completing the square.

Starting from f(x) = -4x² -20x +56.

First, factor out the coefficient of x² from the first two terms:

f(x) = -4(x² + 5x) + 56.

Now, to complete the square inside the parentheses. The coefficient of x is 5. Take half of 5, which is 2.5, and square it, which is 6.25.

So, add and subtract this inside the parentheses:

f(x) = -4[ (x² + 5x + 6.25 - 6.25) ] + 56

This becomes:

-4[ (x + 2.5)^2 - 6.25 ] + 56

Distribute the -4:

-4(x + 2.5)^2 + 25 + 56

Combine constants:

25 + 56 = 81

Therefore, f(x) = -4(x + 2.5)^2 + 81.

So, the maximum value is 81, occurring at x = -2.5. Therefore, that confirms my earlier calculation.

Alternatively, maybe I can use calculus. Taking the derivative of f(x) and setting it to zero.

But since this is a quadratic, calculus might be overkill, but let me try it for verification.

f(x) = -4x² -20x +56

f'(x) = -8x -20

Set derivative equal to zero:

-8x -20 = 0

-8x = 20

x = 20 / (-8) = -2.5

Same result. Then plug back into f(x), which gives 81 as before.

So, all methods lead to 81 as the maximum value.

Alternatively, maybe I can think of the original expression without expanding. The original function is 4(x + 7)(2 - x). Let me think if there's another way to find the maximum value.

Notice that (x + 7)(2 - x) can be rewritten as -(x + 7)(x - 2). Which is similar to a quadratic in terms of x. But since we already have that, maybe another approach is to think of it as a product of two linear terms and find its maximum.

Alternatively, use the vertex formula for a quadratic in factored form. Wait, but the standard vertex formula is for the standard form. However, if I can write the function in the form a(x - h)^2 + k, then k is the maximum or minimum.

But since we already did that earlier, it's the same as completing the square.

Alternatively, since it's a quadratic, the maximum occurs at the vertex, so x = -b/(2a) as we did earlier.

Alternatively, maybe using substitution. Let me think. Let me consider variables. Let me set y = x + 7 and z = 2 - x. Then, the product is 4yz. But I need to maximize yz. However, y and z are related because y = x + 7 and z = 2 - x. Therefore, adding them together: y + z = x + 7 + 2 - x = 9. So, y + z = 9.

Therefore, we need to maximize the product yz given that y + z = 9.

This is a classic optimization problem where for two numbers with a fixed sum, their product is maximized when they are equal. Wait, but that would be if they are positive. However, here y and z could be positive or negative? Let me check.

But since y = x + 7 and z = 2 - x, depending on the value of x, y and z can be positive or negative. However, if we want to maximize yz, given that y + z = 9.

But the maximum of yz when y + z is constant is when y = z = 4.5. However, is that possible? Let me check.

If y = z = 4.5, then y + z = 9, which is correct. Then, y = x + 7 = 4.5 => x = 4.5 - 7 = -2.5. Similarly, z = 2 - x = 4.5 => x = 2 - 4.5 = -2.5. So, that's consistent. Therefore, the maximum occurs at x = -2.5, which gives y = z = 4.5. Therefore, the maximum product yz is (4.5)^2 = 20.25. Then, multiplying by 4 gives 4 * 20.25 = 81. Which again gives 81.

So, this method also gives the same result. So that's a nice check.

Therefore, all methods agree that the maximum value is 81.

Alternatively, maybe I can use a graphing approach. If I imagine the graph of the function f(x) = 4(x + 7)(2 - x), which is a parabola opening downward, since the coefficient of x² is negative. The roots of the function are at x = -7 and x = 2. So, the parabola crosses the x-axis at -7 and 2, and opens downward. The vertex is halfway between the roots? Wait, no, the vertex is at the midpoint between the roots only if the parabola is symmetric. But since this is a standard quadratic, the axis of symmetry is halfway between the roots.

Wait, the roots are at x = -7 and x = 2. The midpoint between -7 and 2 is (-7 + 2)/2 = (-5)/2 = -2.5, which is the same x-coordinate as the vertex we found earlier. Therefore, the vertex is at x = -2.5, which is exactly halfway between the roots. Therefore, the maximum value is at x = -2.5, and substituting back gives 81, as we found.

Therefore, all methods confirm that the maximum value is 81.

But let me confirm once again by plugging in some values around x = -2.5 to see whether the function is indeed maximum there.

For example, let's take x = -2.5, which gives f(x) = 81.

Take x = -3:

f(-3) = 4(-3 + 7)(2 - (-3)) = 4(4)(5) = 4*20 = 80.

Which is less than 81.

Take x = -2:

f(-2) = 4(-2 + 7)(2 - (-2)) = 4(5)(4) = 4*20 = 80.

Also 80, less than 81.

Take x = -2.5:

f(-2.5) = 81.

Take x = -2.6:

f(-2.6) = 4(-2.6 + 7)(2 - (-2.6)) = 4(4.4)(4.6). Let me calculate that:

4 * 4.4 * 4.6. Let's compute 4.4 * 4.6 first.

Note that 4.4 * 4.6 = (4 + 0.4)(4 + 0.6) = 4*4 + 4*0.6 + 0.4*4 + 0.4*0.6 = 16 + 2.4 + 1.6 + 0.24 = 16 + 4 + 0.24 = 20.24.

Then multiply by 4: 20.24 * 4 = 80.96. Which is approximately 81, but slightly less.

Similarly, take x = -2.4:

f(-2.4) = 4(-2.4 + 7)(2 - (-2.4)) = 4(4.6)(4.4). Which is same as above, 4 * 4.4 * 4.6 = 80.96.

Therefore, confirming that at x = -2.5, the value is 81, which is indeed higher than the values at x = -2.6 and x = -2.4.

Therefore, this numerical check also supports that the maximum is 81.

Alternatively, if I take x = -2.5 + h and compute f(x) for small h, the function should have a maximum there.

But given that all methods lead to the same answer, and multiple checks confirm it, I can be confident that the maximum value is 81.

So, the answer is 81.

**Final Answer**
The maximum value is \boxed{81}.

To determine the maximum value of the expression
$$
4(x + 7)(2 - x)
$$
over all real numbers $ x $, we proceed by analyzing and simplifying the expression.

---

### Step 1: Expand the Product

We first simplify the expression:
$$
4(x + 7)(2 - x)
$$

Expanding the binomial:
$$
(x + 7)(2 - x) = -x^2 - 5x + 14
$$

Now multiply by 4:
$$
4(-x^2 - 5x + 14) = -4x^2 - 20x + 56
$$

So, the function becomes:
$$
f(x) = -4x^2 - 20x + 56
$$

This is a **quadratic function** with a **negative leading coefficient** ($-4$), which means the parabola opens **downward**. Therefore, the function has a **maximum value** at its **vertex**.

---

### Step 2: Find the Vertex

For a quadratic of the form $ f(x) = ax^2 + bx + c $, the x-coordinate of the vertex is given by:
$$
x = -\frac{b}{2a}
$$

Here, $ a = -4 $ and $ b = -20 $, so:
$$
x = -\frac{-20}{2(-4)} = \frac{20}{-8} = -2.5
$$

---

### Step 3: Evaluate the Function at the Vertex

To find the maximum value, evaluate $ f(-2.5) $:
$$
f(-2.5) = -4(-2.5)^2 - 20(-2.5) + 56
$$

Compute step-by-step:
- $ (-2.5)^2 = 6.25 $
- $ -4(6.25) = -25 $
- $ -20(-2.5) = 50 $
- So:
$$
f(-2.5) = -25 + 50 + 56 = 81
$$

---

### Step 4: Verify by Completing the Square (Optional)

We can verify this result by rewriting:
$$
f(x) = -4x^2 - 20x + 56
$$
as:
$$
f(x) = -4(x^2 + 5x) + 56
$$
Complete the square inside the parentheses:
$$
x^2 + 5x = (x + 2.5)^2 - 6.25
$$
So:
$$
f(x) = -4\left[(x + 2.5)^2 - 6.25\right] + 56 = -4(x + 2.5)^2 + 25 + 56 = -4(x + 2.5)^2 + 81
$$
This confirms the function achieves its **maximum value of 81** at $ x = -2.5 $, since the square term is always non-positive and minimized when it equals zero.

---

### Final Answer

The maximum value of the function is:

$$
\boxed{81}
$$

Solution:
The graph of $y = 4(x + 7)(2 - x)$ is a parabola.  Since $y = 0$ when $x = -7$ and $x = 2$, the $x$-intercepts of the parabola are $(-7,0)$ and $(2,0)$.  If the vertex of the parabola is $(h,k)$, then the $x$-intercepts $(-7,0)$ and $(2,0)$ are symmetric around the line $x = h$, so $h = (-7 + 2)/2 = -5/2$.

Hence, the maximum value of $y = 4(x + 7)(2 - x)$ occurs at $x = -5/2$, in which case \[y = 4 \left( -\frac{5}{2} + 7 \right) \left( 2 + \frac{5}{2} \right) = 4 \cdot \frac{9}{2} \cdot \frac{9}{2} = \boxed{81}.\] (Note that this is a maximum value, and not a minimum value, because the coefficient of $x^2$ in $y = 4(x + 7)(2 - x) = -4x^2 - 20x + 56$ is negative.)


Explanation:
Okay, so Pat wants to select six cookies from a tray that has chocolate chip, oatmeal, and peanut butter cookies. And there are at least six of each type, so we don't have to worry about running out of any particular kind. The question is asking how many different assortments of six cookies can be selected. Also, it mentions that cookies of the same type aren't distinguishable.

Hmm, let me think. This sounds like a combinatorics problem. Specifically, I remember something about combinations with repetition. Maybe it's similar to the problem where you want to distribute identical items into different boxes? Or maybe it's like finding the number of non-negative integer solutions to an equation?

Let me recall. If we have to select six cookies, and there are three types, then we need to figure out how many different ways we can have combinations of these three types that add up to six. For example, maybe 3 chocolate chip, 2 oatmeal, and 1 peanut butter. Or 6 chocolate chip and none of the others. Each different combination is an "assortment".

So, translating that into math terms, let me denote the number of chocolate chip cookies as x, oatmeal as y, and peanut butter as z. Then the problem becomes finding the number of non-negative integer solutions to the equation:

x + y + z = 6

Where x, y, z ≥ 0.

Yes, that seems right. So, the question reduces to finding the number of non-negative integer solutions to this equation. I think there is a formula for this. Isn't it the stars and bars theorem?

Let me remember. The formula for the number of non-negative integer solutions to x1 + x2 + ... + xk = n is C(n + k - 1, k - 1). Wait, or is it C(n + k - 1, n)? Let me check.

Suppose we have n identical items and k distinct boxes. The number of ways is C(n + k - 1, k - 1). Alternatively, it's also equal to C(n + k - 1, n). Since C(a, b) = C(a, a - b). So both expressions are equivalent.

In this case, n is 6 (the number of cookies) and k is 3 (the types of cookies). Therefore, the number of solutions should be C(6 + 3 - 1, 3 - 1) = C(8, 2). Let me calculate that.

C(8, 2) is 8! / (2! * (8 - 2)!) = (8 * 7) / (2 * 1) = 28.

Wait, so is the answer 28? But let me make sure I didn't mix up the formula. Let me think again.

Alternatively, think of the problem as putting 6 cookies into 3 types. Each type can have 0 or more cookies. So stars and bars. The formula is indeed C(n + k - 1, k - 1). So n is 6, k is 3. So, C(6 + 3 -1, 3 - 1) = C(8, 2) = 28. That seems correct.

But let me try a different approach to verify. Let's consider all possible numbers of each cookie. Since there are three types, we can list all possible combinations. For example:

Case 1: All six cookies are the same type. There are 3 possibilities: all chocolate chip, all oatmeal, or all peanut butter.

Case 2: Five of one type and one of another. How many? Let's see. Choose the type that has five cookies: 3 choices. Then choose the type that has one cookie: 2 choices. So 3 * 2 = 6. But wait, does this count each combination? Yes, since the two types are different. So total 6.

Case 3: Four of one type, one of another, and one of another. So, first choose the type that has four cookies: 3 choices. Then, choose two different types from the remaining two types to have one each. Since there are two remaining types, the number of ways is C(2, 2) = 1. So total for this case: 3 * 1 = 3. Wait, but hold on. Wait, if we have four of one type and one each of the other two, that's a specific distribution. But how many distinct assortments is that? For example, 4 chocolate, 1 oatmeal, 1 peanut butter. Similarly for other types. So there are 3 such assortments.

Case 4: Three of one type, and three of another. So, choose two types to have three each. The number of ways is C(3, 2) = 3. For example, chocolate and oatmeal both have three, chocolate and peanut butter both have three, or oatmeal and peanut butter both have three.

Case 5: Three of one type, two of another, and one of the third. So, first choose which type has three: 3 choices. Then, choose which type has two: 2 choices. Then the remaining type has one. So 3 * 2 = 6.

Case 6: Two of one type, two of another, and two of the third. That's one case. All types have two cookies. So only 1 possibility here.

Case 7: Two of one type, and four of another. Wait, isn't that covered in case 3? Wait, no. Wait, in case 3 we had four of one type and one of each other. Here, two of one type and four of another. Wait, that might be another case. Wait, but when we considered case 2, we had five and one. So maybe I missed some cases?

Wait, let me check. If I list all possible partitions of 6 into 3 parts where order doesn't matter, considering that each part can be zero. Wait, actually, since the cookies are of different types, the order does matter in the sense that which type has how many.

But maybe my initial case breakdown is incomplete. Let me try to think of all the possible integer partitions of 6 into 3 non-negative integers, considering that the order matters (since different types are different). Wait, but actually, the number of solutions is the same as distributing 6 indistinct items into 3 distinct boxes. So the formula is C(8, 2) = 28. So if I try to count by cases, I should get 28. Maybe my earlier cases were missing some.

Alternatively, maybe there's a better way. Let me try to list all possible distributions. Let me consider different partitions of 6 into up to 3 parts. The number of partitions of 6 into at most 3 parts, considering that parts can be zero. Wait, but since we have exactly 3 types, each with a non-negative number of cookies, the problem is equivalent to partitions of 6 into 3 non-negative integers. So, the different ways.

Alternatively, think of the problem as the number of multisets of size 6 from a 3-element set. Which is the same as the formula I mentioned earlier. Which is C(6 + 3 -1, 3 -1) = 28. So maybe the answer is 28, but to make sure, let me count the cases properly.

Alternatively, maybe there's a miscalculation in my first case breakdown. Let me check again.

First, list all possible distributions. Let me think of the possible combinations. Let me think of the number of cookies for each type as (x, y, z), where x + y + z = 6, and x, y, z ≥ 0. The number of such triples is C(8, 2) = 28.

Alternatively, maybe I can compute the number of solutions by considering different scenarios.

For example, the number of solutions where all are different: but since there are three types, maybe not. Wait, perhaps it's better to think in terms of the number of ways to distribute the cookies.

Alternatively, maybe my initial case breakdown was wrong. Let's try to recount:

Case 1: All six of the same type: 3 ways.

Case 2: Five of one type, one of another: For each of the three types, we can have five of that, and one of another. Since there are 3 choices for the type with five cookies, and then 2 choices for the type with one cookie. So 3*2=6.

Case 3: Four of one type, two of another, and zero of the third: Wait, earlier I considered four, one, one. Maybe I need to separate different distributions. Let me think.

Wait, perhaps my earlier cases weren't covering all scenarios. Let me try to categorize based on the number of different types used in the assortment.

Case 1: All six cookies are of one type. So, only one type is used. Number of ways: 3.

Case 2: Two types are used. So, the cookies are divided between two types. For example, 5 and 1, 4 and 2, 3 and 3.

Case 3: All three types are used.

So, let me compute each case.

Case 1: 3 ways.

Case 2: Two types. Let me compute how many ways. For two types, the number of ways to choose the two types is C(3, 2) = 3. For each pair of types, how many ways can we distribute the six cookies between them?

But here, the distribution must be at least one cookie of each type. Wait, no. Wait, in case 2, since we are considering that two types are used, but the problem is that the distribution could be, for example, 6 and 0 (but that would be only one type, not two). Wait, so actually, if we want exactly two types, then we need to have at least one of each. So, for each pair of types, the number of ways is the number of positive integer solutions to x + y = 6, where x and y are at least 1. Which is equivalent to the number of positive integer solutions, which is 5. Wait, let me recall. The formula for positive integer solutions to x + y = n is n - 1. So here, n=6, so 5 solutions. So for each pair of types, there are 5 ways. Therefore, total for case 2: 3 pairs * 5 = 15.

But wait, let me check with an example. For example, take types A and B. The possible distributions are (1,5), (2,4), (3,3), (4,2), (5,1). So 5 ways. That's correct. So case 2 gives 15.

Case 3: All three types are used. So, we need to find the number of solutions where x, y, z ≥ 1 and x + y + z = 6. To compute that, we can use the substitution: let x' = x - 1, y' = y - 1, z' = z - 1. Then x' + y' + z' = 6 - 3 = 3, where x', y', z' ≥ 0. The number of non-negative solutions is C(3 + 3 - 1, 3 - 1) = C(5, 2) = 10. Therefore, case 3: 10 ways.

Therefore, total number of assortments is case1 + case2 + case3 = 3 + 15 + 10 = 28. Which matches the formula result.

So that seems to confirm that the answer is 28.

Alternatively, let me check again the case 2. So, if we think of the case where two types are used, the number is 3 * 5 = 15. However, if we think of the previous case breakdown where I considered five and one, four and two, etc., maybe that's overlapping with other cases. But in this case, since we split into case1, case2, case3, it's comprehensive.

Alternatively, let's think of another way. If I use generating functions. The generating function for each type of cookie is 1 + x + x^2 + x^3 + ... since there's an unlimited supply. And since there are three types, the generating function is (1 + x + x^2 + x^3 + ...)^3. We need the coefficient of x^6 in this expansion.

But since (1 + x + x^2 + ... ) = 1/(1 - x), so the generating function is 1/(1 - x)^3. The coefficient of x^6 in 1/(1 - x)^3 is C(6 + 3 - 1, 3 - 1) = C(8, 2) = 28, same as before.

Therefore, all methods are leading to 28. Therefore, I think the answer is 28. So the number of different assortments is 28.

But let me check once again by enumerating the possible distributions for case 3. Case 3 is when all three types are used. So, each type has at least one cookie. So, the number of distributions is the same as the number of integer solutions to x + y + z = 6, where x, y, z ≥ 1. Which is equivalent to x' + y' + z' = 3, where x', y', z' ≥ 0. Which is C(3 + 3 -1, 3 -1) = C(5, 2) = 10.

Let me think of all possible distributions here. For example:

(1,1,4), (1,4,1), (4,1,1),

(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1),

(2,2,2).

So how many is that?

First, the distributions where one type is 4, and the other two are 1: there are 3 permutations (since which type is 4).

Then, the distributions where one type is 3, another is 2, and the third is 1: there are 3! = 6 permutations.

Then, the distribution where all are 2: only 1 way.

Total: 3 + 6 + 1 = 10. Yes, that's correct. Therefore case3 is 10.

So adding up case1 (3) + case2 (15) + case3 (10) = 28.

Therefore, the answer is 28.

Alternatively, if I didn't remember stars and bars, but I can think of it as combinations with repetition. The formula for combinations with repetition is C(n + k -1, k -1). But here, n is the number of items to choose, which is 6, and k is the number of types, which is 3. Therefore, it's C(6 + 3 -1, 3 -1) = C(8, 2) = 28.

Therefore, regardless of the approach, I get 28.

But just to be thorough, let me check with another approach. Let me use the stars and bars method. Imagine that we have 6 cookies, which we can represent as 6 stars. To divide them into three types, we need two bars to separate them into three groups. For example, **|*|*** would represent 2 chocolate chip, 1 oatmeal, and 3 peanut butter.

The number of ways to arrange these stars and bars is the number of ways to choose positions for the bars. There are 6 stars and 2 bars, so total of 8 symbols. The number of ways is C(8, 2) = 28. Which is the same answer.

Therefore, yes, this confirms it again.

So, all these different approaches—case breakdown, generating functions, stars and bars—lead me to believe that the answer is 28. Therefore, the number of different assortments of six cookies is 28.

**Final Answer**
The number of different assortments of six cookies is \boxed{28}.

To determine how many different assortments of six cookies can be selected from a tray containing chocolate chip, oatmeal, and peanut butter cookies (with at least six of each type available), we can model the problem mathematically.

---

### **Problem Analysis**

We are given:
- A total of **6 cookies** to choose.
- There are **3 types** of cookies: chocolate chip (C), oatmeal (O), and peanut butter (P).
- The cookies of the **same type are indistinguishable** (e.g., all chocolate chip cookies are the same).
- We can choose **any number** of each type, including zero, **as long as the total is 6**.
- The tray has **at least 6 cookies of each type**, so we don’t need to worry about limited quantities.

This is a classic **combinations with repetition** problem, which asks:
> How many ways can we distribute **6 identical items** into **3 distinct categories** (cookie types)?

Let:
- $ x $ = number of chocolate chip cookies
- $ y $ = number of oatmeal cookies
- $ z $ = number of peanut butter cookies

We need to find the number of non-negative integer solutions to the equation:

$$
x + y + z = 6 \quad \text{where } x, y, z \geq 0
$$

---

### **Mathematical Solution**

This is a well-known combinatorics problem that can be solved using the **Stars and Bars** theorem:

- The number of non-negative integer solutions to the equation $ x_1 + x_2 + \dots + x_k = n $ is given by:

$$
\binom{n + k - 1}{k - 1}
$$

Here:
- $ n = 6 $ (total cookies to select)
- $ k = 3 $ (number of types of cookies)

So the number of solutions is:

$$
\binom{6 + 3 - 1}{3 - 1} = \binom{8}{2}
$$

Now calculate:

$$
\binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28
$$

---

### **Conclusion**

There are **28 different assortments** of six cookies that can be selected from the tray, considering the variety of cookie types and that cookies of the same type are indistinguishable.

$$
\boxed{28}
$$

[Assistant]
Explanation:
Okay, so Pat wants to select six cookies from a tray that has chocolate chip, oatmeal, and peanut butter cookies. And there are at least six of each type, so we don't have to worry about running out of any particular kind. The question is asking how many different assortments of six cookies can be selected. Also, it mentions that cookies of the same type aren't distinguishable.

Hmm, let me think. This sounds like a combinatorics problem. Specifically, I remember something about combinations with repetition. Maybe it's similar to the problem where you want to distribute identical items into different boxes? Or maybe it's like finding the number of non-negative integer solutions to an equation?

Let me recall. If we have to select six cookies, and there are three types, then we need to figure out how many different ways we can have combinations of these three types that add up to six. For example, maybe 3 chocolate chip, 2 oatmeal, and 1 peanut butter. Or 6 chocolate chip and none of the others. Each different combination is an "assortment".

So, translating that into math terms, let me denote the number of chocolate chip cookies as x, oatmeal as y, and peanut butter as z. Then the problem becomes finding the number of non-negative integer solutions to the equation:

x + y + z = 6

Where x, y, z ≥ 0.

Yes, that seems right. So, the question reduces to finding the number of non-negative integer solutions to this equation. I think there is a formula for this. Isn't it the stars and bars theorem?

Let me remember. The formula for the number of non-negative integer solutions to x1 + x2 + ... + xk = n is C(n + k - 1, k - 1). Wait, or is it C(n + k - 1, n)? Let me check.

Suppose we have n identical items and k distinct boxes. The number of ways is C(n + k - 1, k - 1). Alternatively, it's also equal to C(n + k - 1, n). Since C(a, b) = C(a, a - b). So both expressions are equivalent.

In this case, n is 6 (the number of cookies) and k is 3 (the types of cookies). Therefore, the number of solutions should be C(6 + 3 - 1, 3 - 1) = C(8, 2). Let me calculate that.

C(8, 2) is 8! / (2! * (8 - 2)!) = (8 * 7) / (2 * 1) = 28.

Wait, so is the answer 28? But let me make sure I didn't mix up the formula. Let me think again.

Alternatively, think of the problem as putting 6 cookies into 3 types. Each type can have 0 or more cookies. So stars and bars. The formula is indeed C(n + k - 1, k - 1). So n is 6, k is 3. So, C(6 + 3 -1, 3 - 1) = C(8, 2) = 28. That seems correct.

But let me try a different approach to verify. Let's consider all possible numbers of each cookie. Since there are three types, we can list all possible combinations. For example:

Case 1: All six cookies are the same type. There are 3 possibilities: all chocolate chip, all oatmeal, or all peanut butter.

Case 2: Five of one type and one of another. How many? Let's see. Choose the type that has five cookies: 3 choices. Then choose the type that has one cookie: 2 choices. So 3 * 2 = 6. But wait, does this count each combination? Yes, since the two types are different. So total 6.

Case 3: Four of one type, one of another, and one of another. So, first choose the type that has four cookies: 3 choices. Then, choose two different types from the remaining two types to have one each. Since there are two remaining types, the number of ways is C(2, 2) = 1. So total for this case: 3 * 1 = 3. Wait, but hold on. Wait, if we have four of one type and one each of the other two, that's a specific distribution. But how many distinct assortments is that? For example, 4 chocolate, 1 oatmeal, 1 peanut butter. Similarly for other types. So there are 3 such assortments.

Case 4: Three of one type, and three of another. So, choose two types to have three each. The number of ways is C(3, 2) = 3. For example, chocolate and oatmeal both have three, chocolate and peanut butter both have three, or oatmeal and peanut butter both have three.

Case 5: Three of one type, two of another, and one of the third. So, first choose which type has three: 3 choices. Then, choose which type has two: 2 choices. Then the remaining type has one. So 3 * 2 = 6.

Case 6: Two of one type, two of another, and two of the third. That's one case. All types have two cookies. So only 1 possibility here.

Case 7: Two of one type, and four of another. Wait, isn't that covered in case 3? Wait, no. Wait, in case 3 we had four of one type and one of each other. Here, two of one type and four of another. Wait, that might be another case. Wait, but when we considered case 2, we had five and one. So maybe I missed some cases?

Wait, let me check. If I list all possible partitions of 6 into 3 parts where order doesn't matter, considering that each part can be zero. Wait, actually, since the cookies are of different types, the order does matter in the sense that which type has how many.

But maybe my initial case breakdown is incomplete. Let me try to think of all the possible integer partitions of 6 into 3 non-negative integers, considering that the order matters (since different types are different). Wait, but actually, the number of solutions is the same as distributing 6 indistinct items into 3 distinct boxes. So the formula is C(8, 2) = 28. So if I try to count by cases, I should get 28. Maybe my earlier cases were missing some.

Alternatively, maybe there's a better way. Let me try to list all possible distributions. Let me consider different partitions of 6 into up to 3 parts. The number of partitions of 6 into at most 3 parts, considering that parts can be zero. Wait, but since we have exactly 3 types, each with a non-negative number of cookies, the problem is equivalent to partitions of 6 into 3 non-negative integers. So, the different ways.

Alternatively, think of the problem as the number of multisets of size 6 from a 3-element set. Which is the same as the formula I mentioned earlier. Which is C(6 + 3 -1, 3 -1) = 28. So maybe the answer is 28, but to make sure, let me count the cases properly.

Alternatively, maybe there's a miscalculation in my first case breakdown. Let me check again.

First, list all possible distributions. Let me think of the possible combinations. Let me think of the number of cookies for each type as (x, y, z), where x + y + z = 6, and x, y, z ≥ 0. The number of such triples is C(8, 2) = 28.

Alternatively, maybe I can compute the number of solutions by considering different scenarios.

For example, the number of solutions where all are different: but since there are three types, maybe not. Wait, perhaps it's better to think in terms of the number of ways to distribute the cookies.

Alternatively, maybe my initial case breakdown was wrong. Let's try to recount:

Case 1: All six of the same type: 3 ways.

Case 2: Five of one type, one of another: For each of the three types, we can have five of that, and one of another. Since there are 3 choices for the type with five cookies, and then 2 choices for the type with one cookie. So 3*2=6.

Case 3: Four of one type, two of another, and zero of the third: Wait, earlier I considered four, one, one. Maybe I need to separate different distributions. Let me think.

Wait, perhaps my earlier cases weren't covering all scenarios. Let me try to categorize based on the number of different types used in the assortment.

Case 1: All six cookies are of one type. So, only one type is used. Number of ways: 3.

Case 2: Two types are used. So, the cookies are divided between two types. For example, 5 and 1, 4 and 2, 3 and 3.

Case 3: All three types are used.

So, let me compute each case.

Case 1: 3 ways.

Case 2: Two types. Let me compute how many ways. For two types, the number of ways to choose the two types is C(3, 2) = 3. For each pair of types, how many ways can we distribute the six cookies between them?

But here, the distribution must be at least one cookie of each type. Wait, no. Wait, in case 2, since we are considering that two types are used, but the problem is that the distribution could be, for example, 6 and 0 (but that would be only one type, not two). Wait, so actually, if we want exactly two types, then we need to have at least one of each. So, for each pair of types, the number of ways is the number of positive integer solutions to x + y = 6, where x and y are at least 1. Which is equivalent to the number of positive integer solutions, which is 5. Wait, let me recall. The formula for positive integer solutions to x + y = n is n - 1. So here, n=6, so 5 solutions. So for each pair of types, there are 5 ways. Therefore, total for case 2: 3 pairs * 5 = 15.

But wait, let me check with an example. For example, take types A and B. The possible distributions are (1,5), (2,4), (3,3), (4,2), (5,1). So 5 ways. That's correct. So case 2 gives 15.

Case 3: All three types are used. So, we need to find the number of solutions where x, y, z ≥ 1 and x + y + z = 6. To compute that, we can use the substitution: let x' = x - 1, y' = y - 1, z' = z - 1. Then x' + y' + z' = 6 - 3 = 3, where x', y', z' ≥ 0. The number of non-negative solutions is C(3 + 3 - 1, 3 - 1) = C(5, 2) = 10. Therefore, case 3: 10 ways.

Therefore, total number of assortments is case1 + case2 + case3 = 3 + 15 + 10 = 28. Which matches the formula result.

So that seems to confirm that the answer is 28.

Alternatively, let me check again the case 2. So, if we think of the case where two types are used, the number is 3 * 5 = 15. However, if we think of the previous case breakdown where I considered five and one, four and two, etc., maybe that's overlapping with other cases. But in this case, since we split into case1, case2, case3, it's comprehensive.

Alternatively, let's think of another way. If I use generating functions. The generating function for each type of cookie is 1 + x + x^2 + x^3 + ... since there's an unlimited supply. And since there are three types, the generating function is (1 + x + x^2 + x^3 + ...)^3. We need the coefficient of x^6 in this expansion.

But since (1 + x + x^2 + ... ) = 1/(1 - x), so the generating function is 1/(1 - x)^3. The coefficient of x^6 in 1/(1 - x)^3 is C(6 + 3 - 1, 3 - 1) = C(8, 2) = 28, same as before.

Therefore, all methods are leading to 28. Therefore, I think the answer is 28. So the number of different assortments is 28.

But let me check once again by enumerating the possible distributions for case 3. Case 3 is when all three types are used. So, each type has at least one cookie. So, the number of distributions is the same as the number of integer solutions to x + y + z = 6, where x, y, z ≥ 1. Which is equivalent to x' + y' + z' = 3, where x', y', z' ≥ 0. Which is C(3 + 3 -1, 3 -1) = C(5, 2) = 10.

Let me think of all possible distributions here. For example:

(1,1,4), (1,4,1), (4,1,1),

(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1),

(2,2,2).

So how many is that?

First, the distributions where one type is 4, and the other two are 1: there are 3 permutations (since which type is 4).

Then, the distributions where one type is 3, another is 2, and the third is 1: there are 3! = 6 permutations.

Then, the distribution where all are 2: only 1 way.

Total: 3 + 6 + 1 = 10. Yes, that's correct. Therefore case3 is 10.

So adding up case1 (3) + case2 (15) + case3 (10) = 28.

Therefore, the answer is 28.

Alternatively, if I didn't remember stars and bars, but I can think of it as combinations with repetition. The formula for combinations with repetition is C(n + k -1, k -1). But here, n is the number of items to choose, which is 6, and k is the number of types, which is 3. Therefore, it's C(6 + 3 -1, 3 -1) = C(8, 2) = 28.

Therefore, regardless of the approach, I get 28.

But just to be thorough, let me check with another approach. Let me use the stars and bars method. Imagine that we have 6 cookies, which we can represent as 6 stars. To divide them into three types, we need two bars to separate them into three groups. For example, **|*|*** would represent 2 chocolate chip, 1 oatmeal, and 3 peanut butter.

The number of ways to arrange these stars and bars is the number of ways to choose positions for the bars. There are 6 stars and 2 bars, so total of 8 symbols. The number of ways is C(8, 2) = 28. Which is the same answer.

Therefore, yes, this confirms it again.

So, all these different approaches—case breakdown, generating functions, stars and bars—lead me to believe that the answer is 28. Therefore, the number of different assortments of six cookies is 28.

**Final Answer**
The number of different assortments of six cookies is \boxed{28}.

To determine how many different assortments of six cookies can be selected from a tray containing chocolate chip, oatmeal, and peanut butter cookies (with at least six of each type available), we can model the problem mathematically.

---

### **Problem Analysis**

We are given:
- A total of **6 cookies** to choose.
- There are **3 types** of cookies: chocolate chip (C), oatmeal (O), and peanut butter (P).
- The cookies of the **same type are indistinguishable** (e.g., all chocolate chip cookies are the same).
- We can choose **any number** of each type, including zero, **as long as the total is 6**.
- The tray has **at least 6 cookies of each type**, so we don’t need to worry about limited quantities.

This is a classic **combinations with repetition** problem, which asks:
> How many ways can we distribute **6 identical items** into **3 distinct categories** (cookie types)?

Let:
- $ x $ = number of chocolate chip cookies
- $ y $ = number of oatmeal cookies
- $ z $ = number of peanut butter cookies

We need to find the number of non-negative integer solutions to the equation:

$$
x + y + z = 6 \quad \text{where } x, y, z \geq 0
$$

---

### **Mathematical Solution**

This is a well-known combinatorics problem that can be solved using the **Stars and Bars** theorem:

- The number of non-negative integer solutions to the equation $ x_1 + x_2 + \dots + x_k = n $ is given by:

$$
\binom{n + k - 1}{k - 1}
$$

Here:
- $ n = 6 $ (total cookies to select)
- $ k = 3 $ (number of types of cookies)

So the number of solutions is:

$$
\binom{6 + 3 - 1}{3 - 1} = \binom{8}{2}
$$

Now calculate:

$$
\binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28
$$

---

### **Conclusion**

There are **28 different assortments** of six cookies that can be selected from the tray, considering the variety of cookie types and that cookies of the same type are indistinguishable.

$$
\boxed{28}
$$

Solution:
The numbers of the three types of cookies must have a sum of six. Possible sets of whole numbers whose sum is six are \[
0,0,6;\ 0,1,5;\ 0,2,4;\ 0,3,3;\ 1,1,4;\ 1,2,3;\ \ \text{and}\ 2,2,2.
\]Every ordering of each of these sets determines a different assortment of cookies. There are 3 orders for each of the sets \[
0,0,6;\ 0,3,3;\ \text{and}\ 1,1,4.
\]There are 6 orders for each of the sets \[
0,1,5;\ 0,2,4;\ \text{and}\ 1,2,3.
\]There is only one order for $2,2,2$. Therefore the total number of assortments of six cookies is $3\cdot 3 + 3\cdot 6 + 1 = \boxed{28}$.