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 unique, and they come in a fixed order. So, the first car is different from the second, etc.

First, let me make sure I understand the problem correctly. We have six cars, each one arriving one after another. Each car can choose to go into one of the three lanes: left, straight, or right. However, we need to ensure that all three lanes are occupied. That means, after all six cars have arrived, each of the three lanes must have at least one car.

So, the problem is similar to assigning each car to one of three lanes, with the constraint that no lane is empty. But since the cars are distinguishable and the order in which they arrive matters, we need to consider permutations with some constraints.

Wait, but actually, the cars are arriving one at a time, so the order in which they arrive is fixed. However, each car can choose which lane to go into. But the problem says "how many ways can the cars stack up so that all three lanes are occupied." So, maybe the stacking up refers to the order in which they are arranged in the lanes?

Wait, but the problem says "the cars are distinguishable, but pull up to the intersection in a fixed order." So, the order of arrival is fixed, but each car can choose which lane to go into. However, the question is about the number of ways the cars can stack up so that all three lanes are occupied.

Wait, maybe I need to think of it as assigning each car to a lane, with the constraint that each lane has at least one car. But since the cars are distinguishable and the order of arrival is fixed, does the order in which they arrive affect the stacking?

Wait, maybe the problem is similar to arranging the cars into the three lanes, considering the order of arrival. For example, if the first car goes into the left lane, the second into straight, etc. But since the cars are distinguishable, the assignment of which car goes into which lane matters.

But the problem says "how many ways can the cars stack up so that all three lanes are occupied." So, perhaps the stacking up refers to the order in which the cars are arranged in each lane? For example, if two cars are in the left lane, their order relative to each other matters?

Wait, but the problem says "the cars pull up to the intersection in a fixed order." So, maybe the order of arrival is fixed, but each car can choose which lane to go into. However, once they choose a lane, they stack up in that lane. So, for example, if the first car goes into the left lane, then the second car could also go into the left lane, and then the first car would be in front of the second car in the left lane.

But the problem says "how many ways can the cars stack up so that all three lanes are occupied." So, perhaps the stacking up refers to the assignment of cars to lanes, considering the order in which they arrive.

Wait, but if the order of arrival is fixed, then the assignment of cars to lanes would be a matter of assigning each car to a lane, with the constraint that all three lanes are non-empty. However, since the cars are distinguishable, the number of ways would be the number of onto functions from the set of cars to the set of lanes, multiplied by something?

Wait, no. Wait, if the order of arrival is fixed, but the assignment to lanes can be done in any way, but we need to count the number of assignments where each lane has at least one car. However, if the order of arrival is fixed, but the stacking up refers to the order in which they are in the lanes.

Wait, maybe the problem is similar to arranging the cars into three lanes, where each lane has an order (since they arrive one at a time, so the order in each lane is determined by the order of arrival).

For example, suppose we have three lanes: left, straight, right. Each car, when it arrives, chooses a lane. Once it chooses a lane, it goes to the back of that lane. So, the order in each lane is the order in which the cars arrived.

But then, the problem is asking for the number of ways the cars can stack up (i.e., assign each car to a lane) such that all three lanes are occupied.

But in that case, the number of ways would be equal to the number of functions from the six cars to the three lanes, minus those functions that don't use all three lanes.

But since the cars are distinguishable, the number of total assignments is 3^6. Then, subtract the assignments where only one or two lanes are used.

But the problem says "stack up so that all three lanes are occupied." So, that would be the inclusion-exclusion principle.

But let me check.

Total number of ways without restriction: 3^6.

Number of ways where at least one lane is empty: ?

Using inclusion-exclusion, the number of onto functions from 6 cars to 3 lanes is 3! * S(6,3), where S(6,3) is the Stirling numbers of the second kind. But since the cars are distinguishable and the lanes are distinguishable, the formula is actually:

Number of onto functions = 3^6 - 3*2^6 + 3*1^6.

Wait, yes, inclusion-exclusion.

So, the formula 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, so it's 3^6 - 3*2^6 + 3*1^6.

Calculating that:

3^6 = 729

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

1^6 = 1, so 3*1 = 3

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

So, 540 ways.

But wait, the problem says "how many ways can the cars stack up so that all three lanes are occupied."

But is this the answer? Wait, but maybe there's more to it.

Wait, the problem says "the cars are distinguishable, but pull up to the intersection in a fixed order." So, does that mean that the order in which they arrive is fixed, but the assignment to lanes can vary?

But if the order of arrival is fixed, then assigning each car to a lane is equivalent to assigning each position in the arrival order to a lane. So, the total number of assignments is 3^6, as before.

But then, the number of assignments where all three lanes are occupied is 540.

But maybe the problem is considering the order in which the cars are arranged in each lane? For example, if two cars are in the left lane, their order in the lane matters?

Wait, but the problem says "the cars pull up to the intersection in a fixed order." So, maybe the order in which they arrive is fixed, so when they go into a lane, they are added to the end of the lane. Therefore, the order in each lane is determined by the arrival order.

But if that's the case, then the assignment of cars to lanes is just a matter of which lane each car goes into, and the order in each lane is determined by the arrival order. Therefore, the number of ways is indeed the number of onto functions, which is 540.

But let me check again.

Alternatively, maybe the problem is considering that the stacking up refers to the order in which the cars are arranged in each lane. For example, if two cars are in the left lane, there are different ways they can be ordered. But since the cars arrive one at a time, the order in each lane is fixed by the arrival order.

For example, if the first car goes into the left lane, and the second car also goes into the left lane, then the first car is in front of the second car in the left lane. So, the order in the lane is determined by the arrival order. Therefore, the assignment of cars to lanes is sufficient to determine the stacking.

Therefore, the total number of ways is the number of assignments of cars to lanes with all three lanes occupied, which is 540.

But let me check with a smaller number. Let's say there are 3 cars and 3 lanes. Then, the number of ways should be 3! = 6.

Using the formula: 3^3 - 3*2^3 + 3*1^3 = 27 - 24 + 3 = 6. Correct.

Another example: 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 = 6*6=36. Correct.

So, for 6 cars, 3 lanes, the number is 540.

But let me think again. The problem says "stack up so that all three lanes are occupied." Maybe there's a different interpretation.

Wait, maybe the problem is not about assigning each car to a lane, but about arranging the cars in the lanes with order. For example, if you have three lanes, each lane can have some number of cars, and the order of cars in each lane matters.

But since the cars arrive one at a time, the order in each lane is determined by the arrival order. So, for example, if car 1 goes to lane A, car 2 goes to lane B, car 3 goes to lane A, then lane A has cars 1 and 3, in that order, and lane B has car 2.

Therefore, the entire configuration is determined by the assignment of each car to a lane. Therefore, the number of such configurations where all three lanes are occupied is equal to the number of onto functions from the 6 cars to the 3 lanes, which is 540.

But maybe the problem is considering that the order in which the cars are arranged in the lanes is different? For example, if two cars are in the same lane, their order could be different? But no, because the cars arrive in a fixed order, so the order in the lane is fixed.

Alternatively, maybe the problem is considering that the stacking up refers to the order in which the cars are arranged in the entire intersection? But that seems less likely.

Alternatively, maybe the problem is about permutations where each car is assigned to a lane, and the order of the cars in the entire queue is considered? But the problem says "the cars pull up to the intersection in a fixed order," so the order of arrival is fixed.

Therefore, I think the answer is 540.

But let me check again.

Alternatively, maybe the problem is considering that each lane has a specific order, and the cars are arranged in the lanes, so that the total number of ways is the number of ways to assign cars to lanes multiplied by the number of orderings within each lane? But no, because the order within each lane is determined by the arrival order.

Wait, suppose that when a car arrives, it goes into a lane, and the order in the lane is the order of arrival. Therefore, once you assign which cars go into which lane, the order in each lane is fixed. Therefore, the total number of configurations is just the number of assignments, which is 3^6.

But if you want all three lanes to be occupied, it's the number of onto functions, which is 540.

Alternatively, if the problem had said that the order in the lanes can be rearranged, then we would have to consider permutations, but since the cars arrive in a fixed order, the order in the lanes is fixed.

Therefore, I think the answer is 540.

But let me check with another approach.

Suppose we think of the problem as first assigning each car to a lane, then ensuring that all three lanes are non-empty.

The number of assignments is 3^6.

Number of assignments where all three lanes are non-empty is equal to the total number of assignments minus those that miss at least one lane.

Using inclusion-exclusion:

Number of assignments missing at least one lane = C(3,1)*2^6 - C(3,2)*1^6 + C(3,3)*0^6.

Therefore, the number of valid assignments is 3^6 - 3*2^6 + 3*1^6 = 729 - 192 + 3 = 540.

Yes, same result.

Alternatively, using Stirling numbers of the second kind. The number of ways to partition 6 distinguishable objects into 3 non-empty subsets is S(6,3). Then, since the lanes are distinguishable, we multiply by 3! to assign each subset to a lane.

So, S(6,3) * 3!

What is S(6,3)?

The formula for Stirling numbers of the second kind is 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, 90 * 6 = 540.

Yes, same answer.

Therefore, the answer is 540.

But let me check S(6,3).

Alternatively, I can compute it.

Stirling numbers of the second kind for S(6,3):

We can use the formula:

S(n,k) = (1/k!) * sum_{i=0 to k} (-1)^{k-i} * C(k,i) * i^n}

But for S(6,3):

S(6,3) = (1/3!)[3^6 - 3*2^6 + 3*1^6]

Which is (1/6)[729 - 192 + 3] = (1/6)(540) = 90.

Yes, so S(6,3) = 90. Then, multiplying by 3! gives 90*6=540.

Therefore, the answer is 540.

But let me check if there's another interpretation.

Suppose that the problem is not about assigning each car to a lane, but about arranging the cars in the three lanes with order. For example, if we think of the entire queue as being divided into three ordered lists (for each lane), then the number of ways would be different.

But in that case, the problem would be similar to arranging the cars into three ordered lists (the lanes), with each list being non-empty.

But since the cars arrive in a fixed order, the order in each lane is determined by the order of arrival. Therefore, the number of ways is equal to the number of ways to assign each car to a lane, with all lanes non-empty, which is again 540.

Alternatively, if the order in the lanes could be arbitrary, then the number would be different. For example, if we had to arrange the cars into three lanes, where the order in each lane matters, and the order of the cars in the entire intersection is not fixed.

But the problem says "the cars pull up to the intersection in a fixed order." So, the order of arrival is fixed, so the order in each lane is determined by the arrival order.

Therefore, the answer is 540.

But let me check with another approach.

Suppose we think of the problem as permutations with constraints.

Each car has three choices, but we need all three lanes to have at least one car.

Alternatively, think of it as the number of sequences of length 6, where each element is L, S, or R, and all three letters appear at least once.

Each such sequence corresponds to an assignment of cars to lanes, with the first car being assigned to the first element, etc.

Therefore, the number of such sequences is equal to the number of onto functions, which is 540.

Yes.

Therefore, I think the answer is 540.

But let me check if the problem says "stack up so that all three lanes are occupied." Maybe "stack up" implies that the order in the lanes matters?

Wait, if the cars are stacking up in the lanes, then maybe the order in which they are in the lane matters. But since they arrive one at a time, the order in the lane is determined by the arrival order.

Alternatively, if the problem had said that the cars can be arranged in any order in the lanes, then we would have to consider permutations.

But given the problem says "the cars pull up to the intersection in a fixed order," I think that the order in which they arrive is fixed, so the order in the lanes is determined by the arrival order.

Therefore, the answer is 540.

But let me check with an example.

Suppose there are 3 cars and 3 lanes. Then, the number of ways should be 3! = 6.

Each car goes into a different lane.

Using the formula, 3^3 - 3*2^3 + 3*1^3 = 27 - 24 + 3 = 6. Correct.

Another example: 4 cars, 3 lanes.

Number of ways is 3^4 - 3*2^4 + 3*1^4 = 81 - 48 + 3 = 36.

Alternatively, think of it as assigning each car to a lane, with all three lanes occupied.

For 4 cars, the possible distributions are 2-1-1, 1-2-1, 1-1-2.

For each distribution, the number of assignments is C(4,2)*C(2,1)*C(1,1) * number of ways to assign the groups to lanes.

Wait, for the distribution 2-1-1:

First, choose which lane has 2 cars: 3 choices.

Then, choose 2 cars out of 4 to assign to that lane: C(4,2).

Then, assign the remaining 2 cars to the other two lanes: 2! ways.

Therefore, total for 2-1-1 is 3 * C(4,2) * 2! = 3 * 6 * 2 = 36.

Which matches the previous result.

Therefore, for 6 cars, the number of ways would be calculated similarly.

But since the formula gives 540, and the inclusion-exclusion approach is correct, I think that is the answer.

Therefore, I think the answer is 540.

But let me check once more.

Alternatively, suppose that the problem is considering that the order in which the cars are in the lanes is important. For example, if two cars are in the same lane, their order in the lane matters. However, since the cars arrive in a fixed order, the order in the lane is determined by the arrival order. Therefore, the assignment of cars to lanes is sufficient to determine the entire configuration.

Therefore, the answer is indeed 540.

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

**Final Answer**
\boxed{540}

To solve the problem, we are given six distinguishable cars arriving one at a time to a red light with three lanes: **left-turn**, **straight-going**, and **right-turn**. The key constraint is that **all three lanes must be occupied** after all six cars have arrived. The cars are distinguishable, and the **order of arrival is fixed**.

---

### Step 1: Understanding the Problem

Each car has **three choices** of lanes: left, straight, or right. Since the cars are distinguishable, the **assignment of each car to a lane** matters. However, the **order in which the cars arrive is fixed**, so the **order of cars within each lane is determined by their arrival order**. That is, if car 1 goes to the left lane and car 2 also goes to the left lane, then car 1 will be in front of car 2 in that lane.

Thus, the **entire configuration** is determined by the **assignment of each car to a lane**, and the **order in each lane is fixed** by the arrival order.

---

### Step 2: Total Assignments Without Restriction

Each of the 6 cars can independently choose one of the 3 lanes. So, the total number of possible assignments (without any restriction) is:

$$
3^6 = 729
$$

---

### Step 3: Applying the Constraint (All Lanes Must Be Occupied)

We now want to count only those assignments where **all three lanes are occupied**. That is, **no lane is empty**.

This is a classic **inclusion-exclusion** problem. We subtract the number of assignments where **at least one lane is empty** from the total.

Let’s define:

- $ A $: assignments where the **left** lane is empty
- $ B $: assignments where the **straight** lane is empty
- $ C $: assignments where the **right** lane is empty

We want to compute:

$$
\text{Valid assignments} = \text{Total} - (|A| + |B| + |C|) + (|A \cap B| + |A \cap C| + |B \cap C|) - |A \cap B \cap C|
$$

Using the principle of inclusion-exclusion:

$$
\text{Valid} = 3^6 - 3 \cdot 2^6 + 3 \cdot 1^6
$$

Now compute each term:

- $ 3^6 = 729 $
- $ 2^6 = 64 $, so $ 3 \cdot 64 = 192 $
- $ 1^6 = 1 $, so $ 3 \cdot 1 = 3 $

$$
\text{Valid} = 729 - 192 + 3 = 540
$$

---

### Step 4: Interpretation

This result, **540**, represents the number of **distinct ways** the six distinguishable cars can be assigned to the three lanes such that **each lane is occupied**. Since the order of arrival is fixed, the **order of cars within each lane is determined** by the arrival order, and thus the **entire configuration is uniquely determined** by the assignment of cars to lanes.

---

### Final Answer

$$
\boxed{540}
$$

[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 unique, and they come in a fixed order. So, the first car is different from the second, etc.

First, let me make sure I understand the problem correctly. We have six cars, each one arriving one after another. Each car can choose to go into one of the three lanes: left, straight, or right. However, we need to ensure that all three lanes are occupied. That means, after all six cars have arrived, each of the three lanes must have at least one car.

So, the problem is similar to assigning each car to one of three lanes, with the constraint that no lane is empty. But since the cars are distinguishable and the order in which they arrive matters, we need to consider permutations with some constraints.

Wait, but actually, the cars are arriving one at a time, so the order in which they arrive is fixed. However, each car can choose which lane to go into. But the problem says "how many ways can the cars stack up so that all three lanes are occupied." So, maybe the stacking up refers to the order in which they are arranged in the lanes?

Wait, but the problem says "the cars are distinguishable, but pull up to the intersection in a fixed order." So, the order of arrival is fixed, but each car can choose which lane to go into. However, the question is about the number of ways the cars can stack up so that all three lanes are occupied.

Wait, maybe I need to think of it as assigning each car to a lane, with the constraint that each lane has at least one car. But since the cars are distinguishable and the order of arrival is fixed, does the order in which they arrive affect the stacking?

Wait, maybe the problem is similar to arranging the cars into the three lanes, considering the order of arrival. For example, if the first car goes into the left lane, the second into straight, etc. But since the cars are distinguishable, the assignment of which car goes into which lane matters.

But the problem says "how many ways can the cars stack up so that all three lanes are occupied." So, perhaps the stacking up refers to the order in which the cars are arranged in each lane? For example, if two cars are in the left lane, their order relative to each other matters?

Wait, but the problem says "the cars pull up to the intersection in a fixed order." So, maybe the order of arrival is fixed, but each car can choose which lane to go into. However, once they choose a lane, they stack up in that lane. So, for example, if the first car goes into the left lane, then the second car could also go into the left lane, and then the first car would be in front of the second car in the left lane.

But the problem says "how many ways can the cars stack up so that all three lanes are occupied." So, perhaps the stacking up refers to the assignment of cars to lanes, considering the order in which they arrive.

Wait, but if the order of arrival is fixed, then the assignment of cars to lanes would be a matter of assigning each car to a lane, with the constraint that all three lanes are non-empty. However, since the cars are distinguishable, the number of ways would be the number of onto functions from the set of cars to the set of lanes, multiplied by something?

Wait, no. Wait, if the order of arrival is fixed, but the assignment to lanes can be done in any way, but we need to count the number of assignments where each lane has at least one car. However, if the order of arrival is fixed, but the stacking up refers to the order in which they are in the lanes.

Wait, maybe the problem is similar to arranging the cars into three lanes, where each lane has an order (since they arrive one at a time, so the order in each lane is determined by the order of arrival).

For example, suppose we have three lanes: left, straight, right. Each car, when it arrives, chooses a lane. Once it chooses a lane, it goes to the back of that lane. So, the order in each lane is the order in which the cars arrived.

But then, the problem is asking for the number of ways the cars can stack up (i.e., assign each car to a lane) such that all three lanes are occupied.

But in that case, the number of ways would be equal to the number of functions from the six cars to the three lanes, minus those functions that don't use all three lanes.

But since the cars are distinguishable, the number of total assignments is 3^6. Then, subtract the assignments where only one or two lanes are used.

But the problem says "stack up so that all three lanes are occupied." So, that would be the inclusion-exclusion principle.

But let me check.

Total number of ways without restriction: 3^6.

Number of ways where at least one lane is empty: ?

Using inclusion-exclusion, the number of onto functions from 6 cars to 3 lanes is 3! * S(6,3), where S(6,3) is the Stirling numbers of the second kind. But since the cars are distinguishable and the lanes are distinguishable, the formula is actually:

Number of onto functions = 3^6 - 3*2^6 + 3*1^6.

Wait, yes, inclusion-exclusion.

So, the formula 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, so it's 3^6 - 3*2^6 + 3*1^6.

Calculating that:

3^6 = 729

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

1^6 = 1, so 3*1 = 3

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

So, 540 ways.

But wait, the problem says "how many ways can the cars stack up so that all three lanes are occupied."

But is this the answer? Wait, but maybe there's more to it.

Wait, the problem says "the cars are distinguishable, but pull up to the intersection in a fixed order." So, does that mean that the order in which they arrive is fixed, but the assignment to lanes can vary?

But if the order of arrival is fixed, then assigning each car to a lane is equivalent to assigning each position in the arrival order to a lane. So, the total number of assignments is 3^6, as before.

But then, the number of assignments where all three lanes are occupied is 540.

But maybe the problem is considering the order in which the cars are arranged in each lane? For example, if two cars are in the left lane, their order in the lane matters?

Wait, but the problem says "the cars pull up to the intersection in a fixed order." So, maybe the order in which they arrive is fixed, so when they go into a lane, they are added to the end of the lane. Therefore, the order in each lane is determined by the arrival order.

But if that's the case, then the assignment of cars to lanes is just a matter of which lane each car goes into, and the order in each lane is determined by the arrival order. Therefore, the number of ways is indeed the number of onto functions, which is 540.

But let me check again.

Alternatively, maybe the problem is considering that the stacking up refers to the order in which the cars are arranged in each lane. For example, if two cars are in the left lane, there are different ways they can be ordered. But since the cars arrive one at a time, the order in each lane is fixed by the arrival order.

For example, if the first car goes into the left lane, and the second car also goes into the left lane, then the first car is in front of the second car in the left lane. So, the order in the lane is determined by the arrival order. Therefore, the assignment of cars to lanes is sufficient to determine the stacking.

Therefore, the total number of ways is the number of assignments of cars to lanes with all three lanes occupied, which is 540.

But let me check with a smaller number. Let's say there are 3 cars and 3 lanes. Then, the number of ways should be 3! = 6.

Using the formula: 3^3 - 3*2^3 + 3*1^3 = 27 - 24 + 3 = 6. Correct.

Another example: 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 = 6*6=36. Correct.

So, for 6 cars, 3 lanes, the number is 540.

But let me think again. The problem says "stack up so that all three lanes are occupied." Maybe there's a different interpretation.

Wait, maybe the problem is not about assigning each car to a lane, but about arranging the cars in the lanes with order. For example, if you have three lanes, each lane can have some number of cars, and the order of cars in each lane matters.

But since the cars arrive one at a time, the order in each lane is determined by the arrival order. So, for example, if car 1 goes to lane A, car 2 goes to lane B, car 3 goes to lane A, then lane A has cars 1 and 3, in that order, and lane B has car 2.

Therefore, the entire configuration is determined by the assignment of each car to a lane. Therefore, the number of such configurations where all three lanes are occupied is equal to the number of onto functions from the 6 cars to the 3 lanes, which is 540.

But maybe the problem is considering that the order in which the cars are arranged in the lanes is different? For example, if two cars are in the same lane, their order could be different? But no, because the cars arrive in a fixed order, so the order in the lane is fixed.

Alternatively, maybe the problem is considering that the stacking up refers to the order in which the cars are arranged in the entire intersection? But that seems less likely.

Alternatively, maybe the problem is about permutations where each car is assigned to a lane, and the order of the cars in the entire queue is considered? But the problem says "the cars pull up to the intersection in a fixed order," so the order of arrival is fixed.

Therefore, I think the answer is 540.

But let me check again.

Alternatively, maybe the problem is considering that each lane has a specific order, and the cars are arranged in the lanes, so that the total number of ways is the number of ways to assign cars to lanes multiplied by the number of orderings within each lane? But no, because the order within each lane is determined by the arrival order.

Wait, suppose that when a car arrives, it goes into a lane, and the order in the lane is the order of arrival. Therefore, once you assign which cars go into which lane, the order in each lane is fixed. Therefore, the total number of configurations is just the number of assignments, which is 3^6.

But if you want all three lanes to be occupied, it's the number of onto functions, which is 540.

Alternatively, if the problem had said that the order in the lanes can be rearranged, then we would have to consider permutations, but since the cars arrive in a fixed order, the order in the lanes is fixed.

Therefore, I think the answer is 540.

But let me check with another approach.

Suppose we think of the problem as first assigning each car to a lane, then ensuring that all three lanes are non-empty.

The number of assignments is 3^6.

Number of assignments where all three lanes are non-empty is equal to the total number of assignments minus those that miss at least one lane.

Using inclusion-exclusion:

Number of assignments missing at least one lane = C(3,1)*2^6 - C(3,2)*1^6 + C(3,3)*0^6.

Therefore, the number of valid assignments is 3^6 - 3*2^6 + 3*1^6 = 729 - 192 + 3 = 540.

Yes, same result.

Alternatively, using Stirling numbers of the second kind. The number of ways to partition 6 distinguishable objects into 3 non-empty subsets is S(6,3). Then, since the lanes are distinguishable, we multiply by 3! to assign each subset to a lane.

So, S(6,3) * 3!

What is S(6,3)?

The formula for Stirling numbers of the second kind is 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, 90 * 6 = 540.

Yes, same answer.

Therefore, the answer is 540.

But let me check S(6,3).

Alternatively, I can compute it.

Stirling numbers of the second kind for S(6,3):

We can use the formula:

S(n,k) = (1/k!) * sum_{i=0 to k} (-1)^{k-i} * C(k,i) * i^n}

But for S(6,3):

S(6,3) = (1/3!)[3^6 - 3*2^6 + 3*1^6]

Which is (1/6)[729 - 192 + 3] = (1/6)(540) = 90.

Yes, so S(6,3) = 90. Then, multiplying by 3! gives 90*6=540.

Therefore, the answer is 540.

But let me check if there's another interpretation.

Suppose that the problem is not about assigning each car to a lane, but about arranging the cars in the three lanes with order. For example, if we think of the entire queue as being divided into three ordered lists (for each lane), then the number of ways would be different.

But in that case, the problem would be similar to arranging the cars into three ordered lists (the lanes), with each list being non-empty.

But since the cars arrive in a fixed order, the order in each lane is determined by the order of arrival. Therefore, the number of ways is equal to the number of ways to assign each car to a lane, with all lanes non-empty, which is again 540.

Alternatively, if the order in the lanes could be arbitrary, then the number would be different. For example, if we had to arrange the cars into three lanes, where the order in each lane matters, and the order of the cars in the entire intersection is not fixed.

But the problem says "the cars pull up to the intersection in a fixed order." So, the order of arrival is fixed, so the order in each lane is determined by the arrival order.

Therefore, the answer is 540.

But let me check with another approach.

Suppose we think of the problem as permutations with constraints.

Each car has three choices, but we need all three lanes to have at least one car.

Alternatively, think of it as the number of sequences of length 6, where each element is L, S, or R, and all three letters appear at least once.

Each such sequence corresponds to an assignment of cars to lanes, with the first car being assigned to the first element, etc.

Therefore, the number of such sequences is equal to the number of onto functions, which is 540.

Yes.

Therefore, I think the answer is 540.

But let me check if the problem says "stack up so that all three lanes are occupied." Maybe "stack up" implies that the order in the lanes matters?

Wait, if the cars are stacking up in the lanes, then maybe the order in which they are in the lane matters. But since they arrive one at a time, the order in the lane is determined by the arrival order.

Alternatively, if the problem had said that the cars can be arranged in any order in the lanes, then we would have to consider permutations.

But given the problem says "the cars pull up to the intersection in a fixed order," I think that the order in which they arrive is fixed, so the order in the lanes is determined by the arrival order.

Therefore, the answer is 540.

But let me check with an example.

Suppose there are 3 cars and 3 lanes. Then, the number of ways should be 3! = 6.

Each car goes into a different lane.

Using the formula, 3^3 - 3*2^3 + 3*1^3 = 27 - 24 + 3 = 6. Correct.

Another example: 4 cars, 3 lanes.

Number of ways is 3^4 - 3*2^4 + 3*1^4 = 81 - 48 + 3 = 36.

Alternatively, think of it as assigning each car to a lane, with all three lanes occupied.

For 4 cars, the possible distributions are 2-1-1, 1-2-1, 1-1-2.

For each distribution, the number of assignments is C(4,2)*C(2,1)*C(1,1) * number of ways to assign the groups to lanes.

Wait, for the distribution 2-1-1:

First, choose which lane has 2 cars: 3 choices.

Then, choose 2 cars out of 4 to assign to that lane: C(4,2).

Then, assign the remaining 2 cars to the other two lanes: 2! ways.

Therefore, total for 2-1-1 is 3 * C(4,2) * 2! = 3 * 6 * 2 = 36.

Which matches the previous result.

Therefore, for 6 cars, the number of ways would be calculated similarly.

But since the formula gives 540, and the inclusion-exclusion approach is correct, I think that is the answer.

Therefore, I think the answer is 540.

But let me check once more.

Alternatively, suppose that the problem is considering that the order in which the cars are in the lanes is important. For example, if two cars are in the same lane, their order in the lane matters. However, since the cars arrive in a fixed order, the order in the lane is determined by the arrival order. Therefore, the assignment of cars to lanes is sufficient to determine the entire configuration.

Therefore, the answer is indeed 540.

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

**Final Answer**
\boxed{540}

To solve the problem, we are given six distinguishable cars arriving one at a time to a red light with three lanes: **left-turn**, **straight-going**, and **right-turn**. The key constraint is that **all three lanes must be occupied** after all six cars have arrived. The cars are distinguishable, and the **order of arrival is fixed**.

---

### Step 1: Understanding the Problem

Each car has **three choices** of lanes: left, straight, or right. Since the cars are distinguishable, the **assignment of each car to a lane** matters. However, the **order in which the cars arrive is fixed**, so the **order of cars within each lane is determined by their arrival order**. That is, if car 1 goes to the left lane and car 2 also goes to the left lane, then car 1 will be in front of car 2 in that lane.

Thus, the **entire configuration** is determined by the **assignment of each car to a lane**, and the **order in each lane is fixed** by the arrival order.

---

### Step 2: Total Assignments Without Restriction

Each of the 6 cars can independently choose one of the 3 lanes. So, the total number of possible assignments (without any restriction) is:

$$
3^6 = 729
$$

---

### Step 3: Applying the Constraint (All Lanes Must Be Occupied)

We now want to count only those assignments where **all three lanes are occupied**. That is, **no lane is empty**.

This is a classic **inclusion-exclusion** problem. We subtract the number of assignments where **at least one lane is empty** from the total.

Let’s define:

- $ A $: assignments where the **left** lane is empty
- $ B $: assignments where the **straight** lane is empty
- $ C $: assignments where the **right** lane is empty

We want to compute:

$$
\text{Valid assignments} = \text{Total} - (|A| + |B| + |C|) + (|A \cap B| + |A \cap C| + |B \cap C|) - |A \cap B \cap C|
$$

Using the principle of inclusion-exclusion:

$$
\text{Valid} = 3^6 - 3 \cdot 2^6 + 3 \cdot 1^6
$$

Now compute each term:

- $ 3^6 = 729 $
- $ 2^6 = 64 $, so $ 3 \cdot 64 = 192 $
- $ 1^6 = 1 $, so $ 3 \cdot 1 = 3 $

$$
\text{Valid} = 729 - 192 + 3 = 540
$$

---

### Step 4: Interpretation

This result, **540**, represents the number of **distinct ways** the six distinguishable cars can be assigned to the three lanes such that **each lane is occupied**. Since the order of arrival is fixed, the **order of cars within each lane is determined** by the arrival order, and thus the **entire configuration is uniquely determined** by the assignment of cars to lanes.

---

### Final Answer

$$
\boxed{540}
$$

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) $. Hmm, let me think about how to approach this. I remember that quadratic functions have either a maximum or a minimum value, depending on the coefficient of the $ x^2 $ term. Since the problem is asking for the maximum value, I should check if this is a quadratic function and then find its vertex.

First, let me expand the expression to make it easier to analyze. Let me multiply out the terms:

$ 4(x + 7)(2 - x) $.

Let me first compute $ (x + 7)(2 - x) $. Let me use the distributive property (FOIL method):

First, multiply $ x \times 2 = 2x $.

Then, $ x \times (-x) = -x^2 $.

Next, $ 7 \times 2 = 14 $.

Finally, $ 7 \times (-x) = -7x $.

So combining all these terms:

$ -x^2 + 2x - 7x + 14 $.

Combine like terms: $ 2x - 7x = -5x $, so the expression becomes:

$ -x^2 - 5x + 14 $.

Now, multiply this by 4:

$ 4(-x^2 - 5x + 14) = -4x^2 - 20x + 56 $.

So the original function simplifies to $ -4x^2 - 20x + 56 $.

Now, since this is a quadratic function in standard form $ ax^2 + bx + c $, where $ a = -4 $, $ b = -20 $, and $ c = 56 $. Since the coefficient $ a $ is negative, the parabola opens downward, which means the vertex will be the maximum point. Therefore, the maximum value occurs at the vertex of this parabola.

The x-coordinate of the vertex of a parabola given by $ ax^2 + bx + c $ is at $ x = -\frac{b}{2a} $. Let me compute that.

Plugging in the values:

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

Wait, let me check that again. The formula is $ -b/(2a) $. Here, $ b = -20 $, so:

$ x = -(-20)/(2 * -4) = 20 / (-8) = -2.5 $. Yes, that's correct. So the x-coordinate of the vertex is -2.5.

Now, to find the maximum value, I need to plug this x-value back into the original function. Let me do that. Alternatively, I can use the vertex formula for the maximum value, which is $ c - \frac{b^2}{4a} $, but I think it's safer to plug in the x-value to avoid mistakes. Let me try both ways to verify.

First, let me plug x = -2.5 into the original function $ 4(x + 7)(2 - x) $.

Calculating $ x + 7 $ when x = -2.5:

$ -2.5 + 7 = 4.5 $.

Calculating $ 2 - x $ when x = -2.5:

$ 2 - (-2.5) = 2 + 2.5 = 4.5 $.

Therefore, the product is $ 4 * 4.5 * 4.5 $. Let me compute that.

First, 4.5 * 4.5 is 20.25. Then, multiplying by 4: 20.25 * 4 = 81. So the maximum value is 81? Wait, that seems straightforward. Let me check with the quadratic form.

Alternatively, using the quadratic function $ -4x^2 - 20x + 56 $, plugging in x = -2.5:

First, compute $ x^2 $: (-2.5)^2 = 6.25.

Then, -4x^2 = -4 * 6.25 = -25.

Next, -20x: -20 * (-2.5) = 50.

Adding the constant term 56: -25 + 50 + 56 = (-25 + 50) + 56 = 25 + 56 = 81. Yep, same result. So that seems correct.

Alternatively, using the vertex formula for maximum value. The formula is $ f(-b/(2a)) $. But another formula for the maximum value is $ \frac{4ac - b^2}{4a} $. Wait, let me recall. The vertex y-coordinate can be calculated as $ c - \frac{b^2}{4a} $? Wait, no, that's not quite right. Let me think.

Actually, the vertex form of a quadratic is $ y = a(x - h)^2 + k $, where (h, k) is the vertex. The maximum or minimum value is k. Alternatively, if you have the standard form, you can compute k by substituting h into the equation.

Alternatively, there's a formula for the maximum value: $ k = c - \frac{b^2}{4a} $? Wait, let me check with the numbers. Let me compute $ c - \frac{b^2}{4a} $:

Here, c = 56, b = -20, a = -4.

So $ 56 - [(-20)^2]/(4*(-4)) = 56 - [400]/(-16) = 56 + 25 = 81. Yes! So that gives the same result. Therefore, the maximum value is indeed 81.

Alternatively, maybe I can use calculus? Taking the derivative and setting it to zero. Let me try that as another method to confirm.

Given the function $ f(x) = 4(x + 7)(2 - x) $. Let me take the derivative f'(x).

First, expand the function again: $ f(x) = -4x^2 -20x + 56 $. Then, derivative is f'(x) = -8x - 20. Setting this equal to zero:

-8x - 20 = 0 => -8x = 20 => x = -20/8 = -2.5. Which is the same x-coordinate as before. Then, plugging back into f(x) gives 81. So same result.

Alternatively, if I didn't expand the function, I could use the product rule. Let me try that. Let me consider f(x) = 4(x + 7)(2 - x). Let me denote u = x + 7 and v = 2 - x. Then f(x) = 4uv. Then, derivative f'(x) = 4(u'v + uv').

Compute u' = derivative of (x + 7) = 1. v' = derivative of (2 - x) = -1. Therefore:

f'(x) = 4[ (1)(2 - x) + (x + 7)(-1) ] = 4[ (2 - x) - (x + 7) ] = 4[ 2 - x - x -7 ] = 4[ -2x -5 ] = -8x -20. Which is the same as before. So setting to zero gives x = -2.5. So again, same result.

Therefore, regardless of the method, the maximum value is 81.

But just to be thorough, let me check with another approach. Maybe completing the square. Let me try that with the quadratic expression.

Starting with $ -4x^2 -20x + 56 $. Let me factor out the coefficient of x^2 from the first two terms:

$ -4(x^2 + 5x) + 56 $.

Now, to complete the square inside the parentheses. The coefficient of x is 5, so half of that is 5/2, and squaring it gives 25/4. Therefore, we add and subtract 25/4 inside the parentheses:

$ -4[ x^2 + 5x + 25/4 - 25/4 ] + 56 = -4[ (x + 5/2)^2 - 25/4 ] + 56 $.

Distribute the -4:

$ -4(x + 5/2)^2 + 4*(25/4) + 56 $.

Simplify:

$ -4(x + 5/2)^2 + 25 + 56 = -4(x + 5/2)^2 + 81 $.

So the function is written as $ -4(x + 5/2)^2 + 81 $. Since the square term is always non-negative, and multiplied by -4, the maximum value occurs when the square term is zero, i.e., when x = -5/2, which is -2.5. At that point, the function equals 81. So again, same result.

Therefore, all methods lead to the conclusion that the maximum value is 81.

But just to make sure, let me test with some numbers. Let me pick x = -2.5, which is -5/2. Let me compute the original expression:

$ 4(x + 7)(2 - x) $.

x + 7 = -5/2 + 7 = (-5/2 + 14/2) = 9/2.

2 - x = 2 - (-5/2) = 2 + 5/2 = 9/2.

Multiply them: (9/2)*(9/2) = 81/4. Multiply by 4: 4*(81/4) = 81. Yep, that's correct.

Let me try another value, say x = 0. Then the expression is 4*(0 + 7)*(2 - 0) = 4*7*2 = 56. Which is less than 81.

How about x = -3? Let's see:

x + 7 = -3 + 7 = 4.

2 - x = 2 - (-3) = 5.

Multiply: 4*5 = 20. Multiply by 4: 80. Close to 81, but less.

x = -2.5 gives 81. Let me try x = -2. Let's see:

x + 7 = -2 + 7 = 5.

2 - x = 2 - (-2) = 4.

Multiply: 5*4 = 20. Multiply by 4: 80. Same as x = -3.

How about x = -2.5 + 0.1 = -2.4?

x + 7 = -2.4 + 7 = 4.6.

2 - x = 2 - (-2.4) = 4.4.

Multiply: 4.6 * 4.4. Let me compute that. 4 * 4.4 = 17.6, 0.6 * 4.4 = 2.64, total is 17.6 + 2.64 = 20.24. Multiply by 4: 80.96, which is slightly less than 81.

Similarly, x = -2.5 - 0.1 = -2.6:

x + 7 = -2.6 + 7 = 4.4.

2 - x = 2 - (-2.6) = 4.6.

Multiply: 4.4 * 4.6 = same as above, 20.24. Multiply by 4: 80.96. So again, less than 81.

Therefore, this numerical check also confirms that the maximum is indeed 81 at x = -2.5.

Alternatively, maybe I can think of the original expression as a quadratic in terms of (x + 7) or something else? Not sure, but since all methods have confirmed the same answer, I think it's safe to conclude that the maximum value is 81.

Wait, just to make sure, let me check if there's any possibility that the maximum occurs at some other point? For example, if the function was not quadratic? But no, expanding the original expression gives a quadratic, so it's a parabola, which has only one maximum or minimum. Since the coefficient is negative, it's a maximum. So there's no other critical points.

Alternatively, if I graph the function, it would be a downward opening parabola with vertex at (-2.5, 81). So the maximum value is 81.

Therefore, after multiple methods and checks, I can confidently say that the maximum value is 81.

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

To find the **maximum value** of the expression:

$$
4(x + 7)(2 - x)
$$

we begin by recognizing that this is a **quadratic expression** in terms of $ x $. The goal is to determine the **maximum value** of this function over all real numbers $ x $.

---

### Step 1: Expand the Expression

We first expand the product:

$$
4(x + 7)(2 - x)
$$

Use the distributive property (FOIL method):

$$
(x + 7)(2 - x) = x(2 - x) + 7(2 - x) = 2x - x^2 + 14 - 7x = -x^2 - 5x + 14
$$

Now multiply by 4:

$$
4(-x^2 - 5x + 14) = -4x^2 - 20x + 56
$$

So the function becomes:

$$
f(x) = -4x^2 - 20x + 56
$$

---

### Step 2: Identify the Type of Function

This is a **quadratic function** of the form:

$$
f(x) = ax^2 + bx + c
$$

with $ a = -4 $, $ b = -20 $, and $ c = 56 $. Since $ a < 0 $, the parabola **opens downward**, meaning the function has a **maximum** at its **vertex**.

---

### Step 3: Find the Vertex

The x-coordinate of the vertex of a parabola is given by:

$$
x = -\frac{b}{2a}
$$

Substitute the values:

$$
x = -\frac{-20}{2(-4)} = \frac{20}{-8} = -2.5
$$

---

### Step 4: Find the Maximum Value

Now substitute $ x = -2.5 $ back into the original expression:

$$
4(x + 7)(2 - x)
$$

Compute:

- $ x + 7 = -2.5 + 7 = 4.5 $
- $ 2 - x = 2 - (-2.5) = 4.5 $

So:

$$
4(4.5)(4.5) = 4 \cdot 20.25 = 81
$$

---

### Step 5: Confirm with Alternative Methods

We can also verify this by completing the square or using calculus:

- **Completing the square** gives:
  $$
  f(x) = -4(x + 2.5)^2 + 81
  $$
  So the maximum value is $ 81 $ when $ x = -2.5 $.

- **Calculus** confirms the same result by taking the derivative and setting it to zero.

---

### Final Answer

$$
\boxed{81}
$$

[Assistant]
Explanation:
Okay, so I need to find the maximum value of the function $ 4(x + 7)(2 - x) $. Hmm, let me think about how to approach this. I remember that quadratic functions have either a maximum or a minimum value, depending on the coefficient of the $ x^2 $ term. Since the problem is asking for the maximum value, I should check if this is a quadratic function and then find its vertex.

First, let me expand the expression to make it easier to analyze. Let me multiply out the terms:

$ 4(x + 7)(2 - x) $.

Let me first compute $ (x + 7)(2 - x) $. Let me use the distributive property (FOIL method):

First, multiply $ x \times 2 = 2x $.

Then, $ x \times (-x) = -x^2 $.

Next, $ 7 \times 2 = 14 $.

Finally, $ 7 \times (-x) = -7x $.

So combining all these terms:

$ -x^2 + 2x - 7x + 14 $.

Combine like terms: $ 2x - 7x = -5x $, so the expression becomes:

$ -x^2 - 5x + 14 $.

Now, multiply this by 4:

$ 4(-x^2 - 5x + 14) = -4x^2 - 20x + 56 $.

So the original function simplifies to $ -4x^2 - 20x + 56 $.

Now, since this is a quadratic function in standard form $ ax^2 + bx + c $, where $ a = -4 $, $ b = -20 $, and $ c = 56 $. Since the coefficient $ a $ is negative, the parabola opens downward, which means the vertex will be the maximum point. Therefore, the maximum value occurs at the vertex of this parabola.

The x-coordinate of the vertex of a parabola given by $ ax^2 + bx + c $ is at $ x = -\frac{b}{2a} $. Let me compute that.

Plugging in the values:

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

Wait, let me check that again. The formula is $ -b/(2a) $. Here, $ b = -20 $, so:

$ x = -(-20)/(2 * -4) = 20 / (-8) = -2.5 $. Yes, that's correct. So the x-coordinate of the vertex is -2.5.

Now, to find the maximum value, I need to plug this x-value back into the original function. Let me do that. Alternatively, I can use the vertex formula for the maximum value, which is $ c - \frac{b^2}{4a} $, but I think it's safer to plug in the x-value to avoid mistakes. Let me try both ways to verify.

First, let me plug x = -2.5 into the original function $ 4(x + 7)(2 - x) $.

Calculating $ x + 7 $ when x = -2.5:

$ -2.5 + 7 = 4.5 $.

Calculating $ 2 - x $ when x = -2.5:

$ 2 - (-2.5) = 2 + 2.5 = 4.5 $.

Therefore, the product is $ 4 * 4.5 * 4.5 $. Let me compute that.

First, 4.5 * 4.5 is 20.25. Then, multiplying by 4: 20.25 * 4 = 81. So the maximum value is 81? Wait, that seems straightforward. Let me check with the quadratic form.

Alternatively, using the quadratic function $ -4x^2 - 20x + 56 $, plugging in x = -2.5:

First, compute $ x^2 $: (-2.5)^2 = 6.25.

Then, -4x^2 = -4 * 6.25 = -25.

Next, -20x: -20 * (-2.5) = 50.

Adding the constant term 56: -25 + 50 + 56 = (-25 + 50) + 56 = 25 + 56 = 81. Yep, same result. So that seems correct.

Alternatively, using the vertex formula for maximum value. The formula is $ f(-b/(2a)) $. But another formula for the maximum value is $ \frac{4ac - b^2}{4a} $. Wait, let me recall. The vertex y-coordinate can be calculated as $ c - \frac{b^2}{4a} $? Wait, no, that's not quite right. Let me think.

Actually, the vertex form of a quadratic is $ y = a(x - h)^2 + k $, where (h, k) is the vertex. The maximum or minimum value is k. Alternatively, if you have the standard form, you can compute k by substituting h into the equation.

Alternatively, there's a formula for the maximum value: $ k = c - \frac{b^2}{4a} $? Wait, let me check with the numbers. Let me compute $ c - \frac{b^2}{4a} $:

Here, c = 56, b = -20, a = -4.

So $ 56 - [(-20)^2]/(4*(-4)) = 56 - [400]/(-16) = 56 + 25 = 81. Yes! So that gives the same result. Therefore, the maximum value is indeed 81.

Alternatively, maybe I can use calculus? Taking the derivative and setting it to zero. Let me try that as another method to confirm.

Given the function $ f(x) = 4(x + 7)(2 - x) $. Let me take the derivative f'(x).

First, expand the function again: $ f(x) = -4x^2 -20x + 56 $. Then, derivative is f'(x) = -8x - 20. Setting this equal to zero:

-8x - 20 = 0 => -8x = 20 => x = -20/8 = -2.5. Which is the same x-coordinate as before. Then, plugging back into f(x) gives 81. So same result.

Alternatively, if I didn't expand the function, I could use the product rule. Let me try that. Let me consider f(x) = 4(x + 7)(2 - x). Let me denote u = x + 7 and v = 2 - x. Then f(x) = 4uv. Then, derivative f'(x) = 4(u'v + uv').

Compute u' = derivative of (x + 7) = 1. v' = derivative of (2 - x) = -1. Therefore:

f'(x) = 4[ (1)(2 - x) + (x + 7)(-1) ] = 4[ (2 - x) - (x + 7) ] = 4[ 2 - x - x -7 ] = 4[ -2x -5 ] = -8x -20. Which is the same as before. So setting to zero gives x = -2.5. So again, same result.

Therefore, regardless of the method, the maximum value is 81.

But just to be thorough, let me check with another approach. Maybe completing the square. Let me try that with the quadratic expression.

Starting with $ -4x^2 -20x + 56 $. Let me factor out the coefficient of x^2 from the first two terms:

$ -4(x^2 + 5x) + 56 $.

Now, to complete the square inside the parentheses. The coefficient of x is 5, so half of that is 5/2, and squaring it gives 25/4. Therefore, we add and subtract 25/4 inside the parentheses:

$ -4[ x^2 + 5x + 25/4 - 25/4 ] + 56 = -4[ (x + 5/2)^2 - 25/4 ] + 56 $.

Distribute the -4:

$ -4(x + 5/2)^2 + 4*(25/4) + 56 $.

Simplify:

$ -4(x + 5/2)^2 + 25 + 56 = -4(x + 5/2)^2 + 81 $.

So the function is written as $ -4(x + 5/2)^2 + 81 $. Since the square term is always non-negative, and multiplied by -4, the maximum value occurs when the square term is zero, i.e., when x = -5/2, which is -2.5. At that point, the function equals 81. So again, same result.

Therefore, all methods lead to the conclusion that the maximum value is 81.

But just to make sure, let me test with some numbers. Let me pick x = -2.5, which is -5/2. Let me compute the original expression:

$ 4(x + 7)(2 - x) $.

x + 7 = -5/2 + 7 = (-5/2 + 14/2) = 9/2.

2 - x = 2 - (-5/2) = 2 + 5/2 = 9/2.

Multiply them: (9/2)*(9/2) = 81/4. Multiply by 4: 4*(81/4) = 81. Yep, that's correct.

Let me try another value, say x = 0. Then the expression is 4*(0 + 7)*(2 - 0) = 4*7*2 = 56. Which is less than 81.

How about x = -3? Let's see:

x + 7 = -3 + 7 = 4.

2 - x = 2 - (-3) = 5.

Multiply: 4*5 = 20. Multiply by 4: 80. Close to 81, but less.

x = -2.5 gives 81. Let me try x = -2. Let's see:

x + 7 = -2 + 7 = 5.

2 - x = 2 - (-2) = 4.

Multiply: 5*4 = 20. Multiply by 4: 80. Same as x = -3.

How about x = -2.5 + 0.1 = -2.4?

x + 7 = -2.4 + 7 = 4.6.

2 - x = 2 - (-2.4) = 4.4.

Multiply: 4.6 * 4.4. Let me compute that. 4 * 4.4 = 17.6, 0.6 * 4.4 = 2.64, total is 17.6 + 2.64 = 20.24. Multiply by 4: 80.96, which is slightly less than 81.

Similarly, x = -2.5 - 0.1 = -2.6:

x + 7 = -2.6 + 7 = 4.4.

2 - x = 2 - (-2.6) = 4.6.

Multiply: 4.4 * 4.6 = same as above, 20.24. Multiply by 4: 80.96. So again, less than 81.

Therefore, this numerical check also confirms that the maximum is indeed 81 at x = -2.5.

Alternatively, maybe I can think of the original expression as a quadratic in terms of (x + 7) or something else? Not sure, but since all methods have confirmed the same answer, I think it's safe to conclude that the maximum value is 81.

Wait, just to make sure, let me check if there's any possibility that the maximum occurs at some other point? For example, if the function was not quadratic? But no, expanding the original expression gives a quadratic, so it's a parabola, which has only one maximum or minimum. Since the coefficient is negative, it's a maximum. So there's no other critical points.

Alternatively, if I graph the function, it would be a downward opening parabola with vertex at (-2.5, 81). So the maximum value is 81.

Therefore, after multiple methods and checks, I can confidently say that the maximum value is 81.

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

To find the **maximum value** of the expression:

$$
4(x + 7)(2 - x)
$$

we begin by recognizing that this is a **quadratic expression** in terms of $ x $. The goal is to determine the **maximum value** of this function over all real numbers $ x $.

---

### Step 1: Expand the Expression

We first expand the product:

$$
4(x + 7)(2 - x)
$$

Use the distributive property (FOIL method):

$$
(x + 7)(2 - x) = x(2 - x) + 7(2 - x) = 2x - x^2 + 14 - 7x = -x^2 - 5x + 14
$$

Now multiply by 4:

$$
4(-x^2 - 5x + 14) = -4x^2 - 20x + 56
$$

So the function becomes:

$$
f(x) = -4x^2 - 20x + 56
$$

---

### Step 2: Identify the Type of Function

This is a **quadratic function** of the form:

$$
f(x) = ax^2 + bx + c
$$

with $ a = -4 $, $ b = -20 $, and $ c = 56 $. Since $ a < 0 $, the parabola **opens downward**, meaning the function has a **maximum** at its **vertex**.

---

### Step 3: Find the Vertex

The x-coordinate of the vertex of a parabola is given by:

$$
x = -\frac{b}{2a}
$$

Substitute the values:

$$
x = -\frac{-20}{2(-4)} = \frac{20}{-8} = -2.5
$$

---

### Step 4: Find the Maximum Value

Now substitute $ x = -2.5 $ back into the original expression:

$$
4(x + 7)(2 - x)
$$

Compute:

- $ x + 7 = -2.5 + 7 = 4.5 $
- $ 2 - x = 2 - (-2.5) = 4.5 $

So:

$$
4(4.5)(4.5) = 4 \cdot 20.25 = 81
$$

---

### Step 5: Confirm with Alternative Methods

We can also verify this by completing the square or using calculus:

- **Completing the square** gives:
  $$
  f(x) = -4(x + 2.5)^2 + 81
  $$
  So the maximum value is $ 81 $ when $ x = -2.5 $.

- **Calculus** confirms the same result by taking the derivative and setting it to zero.

---

### Final Answer

$$
\boxed{81}
$$

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. The problem says there are at least six of each type, so we don't have to worry about running out of any particular type. The question is asking how many different assortments of six cookies can be selected. Also, cookies of the same type aren't distinguishable, which means that the order in which Pat picks them doesn't matter, just the number of each type.

Hmm, so this seems like a combinatorics problem. Specifically, I think it's a stars and bars problem. Let me recall: when you want to distribute n identical items into k distinct categories, the number of ways to do that is given by the combination formula C(n + k - 1, k - 1). Is that right? Let me check with a simple example. If I have 2 cookies and 2 types, then the number of ways should be C(2 + 2 - 1, 2 - 1) = C(3,1) = 3. Which makes sense: both cookies of type 1, both of type 2, or one of each. Yeah, that seems correct.

So applying that formula here, we have n = 6 cookies to select, and k = 3 types of cookies. Therefore, the number of different assortments should be C(6 + 3 - 1, 3 - 1) = C(8, 2). Let me compute that. C(8,2) is 8*7 / 2*1 = 28. So is the answer 28? Wait, but let me make sure I didn't skip any steps or misunderstand the problem.

Wait, the problem says "different assortments of six cookies". Since the cookies of the same type are not distinguishable, the assortment is determined by the number of each type. So, for example, 2 chocolate chip, 3 oatmeal, 1 peanut butter is different from 3 chocolate chip, 2 oatmeal, 1 peanut butter, etc. So yes, this is exactly the problem of finding the number of non-negative integer solutions to the equation:

c + o + p = 6,

where c, o, p are the numbers of chocolate chip, oatmeal, and peanut butter cookies respectively. Each variable is a non-negative integer (since you can have zero of a type), and we need the number of solutions.

So, using stars and bars, the formula is indeed C(n + k - 1, k - 1) where n is the number of items to distribute (6 cookies) and k is the number of categories (3 types). Therefore, substituting, we get C(6 + 3 - 1, 3 - 1) = C(8, 2) = 28. So that seems straightforward.

But let me think again if there are any constraints. The problem says there are at least six of each kind of cookie on the tray. Wait, does that affect anything? Because if there were fewer than six of a certain type, we would have to adjust the formula. But since there are at least six of each, that means that we can have up to six of any type, but since we are only selecting six cookies total, the maximum number of any one type we could have is six. But since the tray has at least six of each, there's no restriction on how many of each type we can take. Therefore, the number of solutions is indeed the same as the number of non-negative integer solutions to c + o + p = 6, which is 28.

Wait, but let me check with another approach. Maybe enumerating some possibilities? Although with 28 possibilities, that's too many, but maybe for smaller numbers.

Alternatively, think of generating functions. The generating function for each type of cookie is 1 + x + x^2 + x^3 + ... since there are unlimited quantities (but in reality, since we are only selecting 6, the terms beyond x^6 won't matter). So 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 + ... ) is 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. So that confirms the same answer. So that seems solid.

Alternatively, maybe think of it as combinations with repetition. The formula for combinations with repetition is C(n + k - 1, k - 1), which is exactly what we used. So that's another way to think about it.

Alternatively, maybe think of the problem as placing dividers between the cookies. For example, if we have 6 cookies, and we want to divide them into 3 types. So imagine 6 stars (cookies) and 2 bars (dividers) to separate them into three groups. The number of ways is C(6 + 2, 2) = C(8, 2) = 28. Yep, that's the same answer.

So all these different approaches lead to the same answer, 28. Therefore, I think that is the correct answer.

But just to be thorough, let me try to think if there's any possible mistake here. For example, sometimes in combinatorics problems, there are constraints that aren't immediately obvious. For example, if the problem had said that Pat must select at least one of each type, then we would have to adjust the formula. But in this case, the problem says "assortments of six cookies", and there's no requirement to have at least one of each type. So, for example, Pat could select all six cookies as chocolate chip, or five chocolate chip and one oatmeal, etc. So the original formula applies.

But let me check with an example. Suppose we have n=2 cookies and k=2 types. Then the number of assortments would be C(2 + 2 -1, 2 -1) = C(3,1)=3. Which are (2,0), (1,1), (0,2). Which is correct. So if we have n=6 and k=3, then the formula gives 28. So that seems correct.

Alternatively, if I try to compute it manually for a smaller number. Let's say n=3, k=3. Then the number of assortments would be C(3 + 3 -1, 3 -1) = C(5,2)=10. Let me list them:

(3,0,0), (0,3,0), (0,0,3),

(2,1,0), (2,0,1), (1,2,0), (0,2,1), (1,0,2), (0,1,2),

(1,1,1).

That's 3 + 6 + 1 = 10. Yep, that works. So the formula is correct.

Therefore, applying the same logic, for n=6, k=3, the answer is 28. Therefore, I think that is the correct answer.

But just to make sure, let me think of another way. Suppose we fix the number of chocolate chip cookies and then compute the possibilities for the remaining.

For example, let c be the number of chocolate chip cookies, which can range from 0 to 6. For each value of c, the number of oatmeal and peanut butter cookies must add up to 6 - c. So for each c, the number of solutions is (6 - c + 2 - 1 choose 2 - 1) = (7 - c choose 1) = 7 - c. Therefore, the total number of solutions is the sum from c=0 to 6 of (7 - c). Which is 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28. Yep, that's the same answer. So that's another way to compute it, and it gives 28. So that's solid.

Therefore, I think the answer is 28. So the number of different assortments 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, we need to consider the nature of the problem:

- **Cookies are indistinguishable within each type** (i.e., all chocolate chip cookies are the same, etc.).
- **Pat is selecting six cookies in total**, and the tray has **at least six of each type**, so there are no restrictions on how many of each type can be selected.
- The goal is to find the number of **distinct combinations** (or "assortments") of six cookies, where the only thing that matters is how many of each type are selected.

---

### Step 1: Model the Problem

Let:
- $ c $ = number of chocolate chip cookies
- $ o $ = number of oatmeal cookies
- $ p $ = number of peanut butter cookies

We are looking for the number of **non-negative integer solutions** to the equation:

$$
c + o + p = 6
$$

This is a classic **stars and bars** problem, where we are distributing 6 identical items (cookies) into 3 distinct categories (types of cookies).

---

### Step 2: Apply the Stars and Bars Formula

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

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

In this case:
- $ n = 6 $ (total cookies)
- $ k = 3 $ (types of cookies)

So the number of solutions is:

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

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

---

### Step 3: Interpret the Result

Each solution corresponds to a unique **assortment** of six cookies, where the count of each type is non-negative and sums to 6. Since the cookies of the same type are indistinguishable, the order of selection does not matter—only the counts of each type.

---

### Final Answer

$$
\boxed{28}
$$

[Assistant]
Explanation:
Okay, so Pat wants to select six cookies from a tray that has chocolate chip, oatmeal, and peanut butter cookies. The problem says there are at least six of each type, so we don't have to worry about running out of any particular type. The question is asking how many different assortments of six cookies can be selected. Also, cookies of the same type aren't distinguishable, which means that the order in which Pat picks them doesn't matter, just the number of each type.

Hmm, so this seems like a combinatorics problem. Specifically, I think it's a stars and bars problem. Let me recall: when you want to distribute n identical items into k distinct categories, the number of ways to do that is given by the combination formula C(n + k - 1, k - 1). Is that right? Let me check with a simple example. If I have 2 cookies and 2 types, then the number of ways should be C(2 + 2 - 1, 2 - 1) = C(3,1) = 3. Which makes sense: both cookies of type 1, both of type 2, or one of each. Yeah, that seems correct.

So applying that formula here, we have n = 6 cookies to select, and k = 3 types of cookies. Therefore, the number of different assortments should be C(6 + 3 - 1, 3 - 1) = C(8, 2). Let me compute that. C(8,2) is 8*7 / 2*1 = 28. So is the answer 28? Wait, but let me make sure I didn't skip any steps or misunderstand the problem.

Wait, the problem says "different assortments of six cookies". Since the cookies of the same type are not distinguishable, the assortment is determined by the number of each type. So, for example, 2 chocolate chip, 3 oatmeal, 1 peanut butter is different from 3 chocolate chip, 2 oatmeal, 1 peanut butter, etc. So yes, this is exactly the problem of finding the number of non-negative integer solutions to the equation:

c + o + p = 6,

where c, o, p are the numbers of chocolate chip, oatmeal, and peanut butter cookies respectively. Each variable is a non-negative integer (since you can have zero of a type), and we need the number of solutions.

So, using stars and bars, the formula is indeed C(n + k - 1, k - 1) where n is the number of items to distribute (6 cookies) and k is the number of categories (3 types). Therefore, substituting, we get C(6 + 3 - 1, 3 - 1) = C(8, 2) = 28. So that seems straightforward.

But let me think again if there are any constraints. The problem says there are at least six of each kind of cookie on the tray. Wait, does that affect anything? Because if there were fewer than six of a certain type, we would have to adjust the formula. But since there are at least six of each, that means that we can have up to six of any type, but since we are only selecting six cookies total, the maximum number of any one type we could have is six. But since the tray has at least six of each, there's no restriction on how many of each type we can take. Therefore, the number of solutions is indeed the same as the number of non-negative integer solutions to c + o + p = 6, which is 28.

Wait, but let me check with another approach. Maybe enumerating some possibilities? Although with 28 possibilities, that's too many, but maybe for smaller numbers.

Alternatively, think of generating functions. The generating function for each type of cookie is 1 + x + x^2 + x^3 + ... since there are unlimited quantities (but in reality, since we are only selecting 6, the terms beyond x^6 won't matter). So 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 + ... ) is 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. So that confirms the same answer. So that seems solid.

Alternatively, maybe think of it as combinations with repetition. The formula for combinations with repetition is C(n + k - 1, k - 1), which is exactly what we used. So that's another way to think about it.

Alternatively, maybe think of the problem as placing dividers between the cookies. For example, if we have 6 cookies, and we want to divide them into 3 types. So imagine 6 stars (cookies) and 2 bars (dividers) to separate them into three groups. The number of ways is C(6 + 2, 2) = C(8, 2) = 28. Yep, that's the same answer.

So all these different approaches lead to the same answer, 28. Therefore, I think that is the correct answer.

But just to be thorough, let me try to think if there's any possible mistake here. For example, sometimes in combinatorics problems, there are constraints that aren't immediately obvious. For example, if the problem had said that Pat must select at least one of each type, then we would have to adjust the formula. But in this case, the problem says "assortments of six cookies", and there's no requirement to have at least one of each type. So, for example, Pat could select all six cookies as chocolate chip, or five chocolate chip and one oatmeal, etc. So the original formula applies.

But let me check with an example. Suppose we have n=2 cookies and k=2 types. Then the number of assortments would be C(2 + 2 -1, 2 -1) = C(3,1)=3. Which are (2,0), (1,1), (0,2). Which is correct. So if we have n=6 and k=3, then the formula gives 28. So that seems correct.

Alternatively, if I try to compute it manually for a smaller number. Let's say n=3, k=3. Then the number of assortments would be C(3 + 3 -1, 3 -1) = C(5,2)=10. Let me list them:

(3,0,0), (0,3,0), (0,0,3),

(2,1,0), (2,0,1), (1,2,0), (0,2,1), (1,0,2), (0,1,2),

(1,1,1).

That's 3 + 6 + 1 = 10. Yep, that works. So the formula is correct.

Therefore, applying the same logic, for n=6, k=3, the answer is 28. Therefore, I think that is the correct answer.

But just to make sure, let me think of another way. Suppose we fix the number of chocolate chip cookies and then compute the possibilities for the remaining.

For example, let c be the number of chocolate chip cookies, which can range from 0 to 6. For each value of c, the number of oatmeal and peanut butter cookies must add up to 6 - c. So for each c, the number of solutions is (6 - c + 2 - 1 choose 2 - 1) = (7 - c choose 1) = 7 - c. Therefore, the total number of solutions is the sum from c=0 to 6 of (7 - c). Which is 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28. Yep, that's the same answer. So that's another way to compute it, and it gives 28. So that's solid.

Therefore, I think the answer is 28. So the number of different assortments 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, we need to consider the nature of the problem:

- **Cookies are indistinguishable within each type** (i.e., all chocolate chip cookies are the same, etc.).
- **Pat is selecting six cookies in total**, and the tray has **at least six of each type**, so there are no restrictions on how many of each type can be selected.
- The goal is to find the number of **distinct combinations** (or "assortments") of six cookies, where the only thing that matters is how many of each type are selected.

---

### Step 1: Model the Problem

Let:
- $ c $ = number of chocolate chip cookies
- $ o $ = number of oatmeal cookies
- $ p $ = number of peanut butter cookies

We are looking for the number of **non-negative integer solutions** to the equation:

$$
c + o + p = 6
$$

This is a classic **stars and bars** problem, where we are distributing 6 identical items (cookies) into 3 distinct categories (types of cookies).

---

### Step 2: Apply the Stars and Bars Formula

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

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

In this case:
- $ n = 6 $ (total cookies)
- $ k = 3 $ (types of cookies)

So the number of solutions is:

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

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

---

### Step 3: Interpret the Result

Each solution corresponds to a unique **assortment** of six cookies, where the count of each type is non-negative and sums to 6. Since the cookies of the same type are indistinguishable, the order of selection does not matter—only the counts of each type.

---

### Final Answer

$$
\boxed{28}
$$

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 $3\cdot 3 + 3\cdot 6 + 1 = \boxed{28}$.