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. And it mentions that cookies of the same type aren't distinguishable.

Alright, let me think. Since the cookies of the same type aren't different, this seems like a problem where I need to count the number of combinations with repetition. Wait, but let me recall the exact terminology here. When you're selecting items where the order doesn't matter and you can have multiple copies of the same item, it's a combination with repetition problem.

But let me make sure. The problem is about selecting six cookies, and the different assortments would be different in terms of how many of each type are selected. For example, one assortment could be 2 chocolate chip, 3 oatmeal, and 1 peanut butter. Another could be 1 chocolate chip, 2 oatmeal, and 3 peanut butter. These are different assortments.

So, the problem reduces to finding the number of non-negative integer solutions to the equation:

c + o + p = 6

where c is the number of chocolate chip cookies, o is the number of oatmeal cookies, and p is the number of peanut butter cookies. Each of c, o, p must be greater than or equal to zero.

Right, so this is a classic stars and bars problem. The formula for the number of non-negative integer solutions to x1 + x2 + ... + xk = n is C(n + k - 1, k - 1). In this case, n is 6 (the total number of cookies to select) and k is 3 (the three types of cookies).

So plugging into the formula, that would be C(6 + 3 - 1, 3 - 1) = C(8, 2). Let me compute that.

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

Wait, so is the answer 28? Let me check if I did that correctly.

Alternatively, maybe I can think of it as arranging stars and bars. If I have 6 cookies, which are stars, and I need to separate them into 3 groups (for the three types). To do that, I need 2 bars. So the total number of symbols is 6 stars and 2 bars, which is 8 symbols. The number of ways to arrange them is C(8, 2), which is 28. That seems right.

But let me think again if there's any catch here. The problem says there are at least six of each type of cookie. Wait, does that affect anything? Because if there were fewer than six of a certain type, we would have to subtract the cases where we take more than available. But since there are at least six of each type, we can take up to six of any one type. But since we are only selecting six cookies total, the maximum number of any one type we could take is six. For example, if we take all six cookies as chocolate chip, that's allowed because there are at least six chocolate chip cookies. Similarly for the others. Therefore, there are no restrictions beyond the non-negative integers.

Therefore, the answer is indeed 28.

But let me try another approach to verify. Let's think about enumerating the possibilities. For example, the number of ways to have 0 chocolate chip cookies: then we have to choose 6 cookies from oatmeal and peanut butter. That would be the number of non-negative solutions to o + p = 6, which is C(6 + 2 - 1, 2 - 1) = C(7, 1) = 7.

Similarly, for 1 chocolate chip cookie: then o + p = 5, which is C(5 + 2 - 1, 2 - 1) = C(6, 1) = 6.

For 2 chocolate chip cookies: o + p = 4, which is C(5, 1) = 5.

Continuing this way:

3 chocolate chip: o + p = 3: C(4, 1) = 4

4 chocolate chip: o + p = 2: C(3, 1) = 3

5 chocolate chip: o + p = 1: C(2, 1) = 2

6 chocolate chip: o + p = 0: C(1, 1) = 1

Now, adding all these up: 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28.

Yes, that's the same answer. So that seems to confirm it.

Alternatively, if I think of the formula for combinations with repetition, which is C(n + k - 1, k - 1). Here n is 6, k is 3, so C(6 + 3 - 1, 3 - 1) = C(8, 2) = 28.

So all methods lead to 28. Therefore, I think the answer is 28.

But just to be thorough, let me think if there's another way someone might approach this problem and perhaps get a different answer. For example, if someone thought that since there are three types, maybe they would think of permutations or something else. But no, since the cookies are not distinguishable except by type, it's purely a matter of how many of each type.

Alternatively, someone might think that since there are three types, the number of assortments is 3^6, but that would be if the cookies were distinguishable, which they are not. So that approach is wrong.

Alternatively, someone might think that since there are three types, the number of ways is similar to arranging letters, but again, that's not the case here.

Alternatively, maybe someone might think that since there are at least six of each type, but since we are only choosing six, maybe there's some restriction? But no, because even if you take all six of one type, there are enough cookies. So no restrictions.

Therefore, I think the answer is 28.

Wait, but let me check with another formula. The number of non-negative integer solutions to c + o + p = 6 is equal to the combination with repetition formula. Which is indeed C(6 + 3 - 1, 3 - 1) = C(8, 2) = 28.

Alternatively, if I use generating functions. The generating function for each type is 1 + x + x^2 + x^3 + ... since there are unlimited cookies (but in our case, since we are only choosing 6, even if there were a limited number, but since there are at least 6, it's okay). So the generating function is (1/(1 - x))^3. We need the coefficient of x^6 in this expansion.

The coefficient of x^n in (1/(1 - x))^k is C(n + k - 1, k - 1). Therefore, coefficient of x^6 is C(6 + 3 - 1, 3 - 1) = C(8, 2) = 28.

Same answer.

Therefore, I think I can be confident that the answer is 28.

But just to make sure, let me think of a smaller case. Suppose instead of 6 cookies, we had 2 cookies, and 3 types. How many assortments?

Using the formula, it would be C(2 + 3 - 1, 3 - 1) = C(4, 2) = 6.

Let me list them:

(0,0,2), (0,1,1), (0,2,0), (1,0,1), (1,1,0), (2,0,0). That's 6.

Yes, that's correct. So the formula works here.

Similarly, if we have 3 cookies, the formula would give C(3 + 3 -1, 3 -1) = C(5, 2) = 10.

Listing them would be tedious, but I can trust the formula.

Therefore, going back to the original problem, 6 cookies, 3 types, answer is 28.

Therefore, I think the answer 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 count the number of **non-negative integer solutions** to the equation:

$$
c + o + p = 6
$$

where:
- $ c $ is the number of chocolate chip cookies,
- $ o $ is the number of oatmeal cookies,
- $ p $ is the number of peanut butter cookies.

Each variable represents a non-negative integer (i.e., $ c, o, p \geq 0 $), and the total number of cookies selected is 6.

---

### Step 1: Recognize the Problem Type

This is a classic **combinations with repetition** problem. We are choosing 6 items (cookies) from 3 types (chocolate chip, oatmeal, peanut butter), and the order of selection does not matter. Since cookies of the same type are indistinguishable, we are counting the number of **distinct combinations** of counts of each type that sum to 6.

This is equivalent to finding the number of **non-negative integer solutions** to the equation:

$$
c + o + p = 6
$$

---

### 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 the **stars and bars** formula:

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

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

So the number of solutions is:

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

---

### Step 3: Compute the Value

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

---

### Step 4: Interpret the Result

Each of these 28 combinations corresponds to a unique **assortment** of six cookies, where the counts of each type of cookie are non-negative integers that sum to 6. Since the problem states there are at least six of each type of cookie, we don't need to worry about exceeding the available quantity of any 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. And it mentions that cookies of the same type aren't distinguishable.

Alright, let me think. Since the cookies of the same type aren't different, this seems like a problem where I need to count the number of combinations with repetition. Wait, but let me recall the exact terminology here. When you're selecting items where the order doesn't matter and you can have multiple copies of the same item, it's a combination with repetition problem.

But let me make sure. The problem is about selecting six cookies, and the different assortments would be different in terms of how many of each type are selected. For example, one assortment could be 2 chocolate chip, 3 oatmeal, and 1 peanut butter. Another could be 1 chocolate chip, 2 oatmeal, and 3 peanut butter. These are different assortments.

So, the problem reduces to finding the number of non-negative integer solutions to the equation:

c + o + p = 6

where c is the number of chocolate chip cookies, o is the number of oatmeal cookies, and p is the number of peanut butter cookies. Each of c, o, p must be greater than or equal to zero.

Right, so this is a classic stars and bars problem. The formula for the number of non-negative integer solutions to x1 + x2 + ... + xk = n is C(n + k - 1, k - 1). In this case, n is 6 (the total number of cookies to select) and k is 3 (the three types of cookies).

So plugging into the formula, that would be C(6 + 3 - 1, 3 - 1) = C(8, 2). Let me compute that.

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

Wait, so is the answer 28? Let me check if I did that correctly.

Alternatively, maybe I can think of it as arranging stars and bars. If I have 6 cookies, which are stars, and I need to separate them into 3 groups (for the three types). To do that, I need 2 bars. So the total number of symbols is 6 stars and 2 bars, which is 8 symbols. The number of ways to arrange them is C(8, 2), which is 28. That seems right.

But let me think again if there's any catch here. The problem says there are at least six of each type of cookie. Wait, does that affect anything? Because if there were fewer than six of a certain type, we would have to subtract the cases where we take more than available. But since there are at least six of each type, we can take up to six of any one type. But since we are only selecting six cookies total, the maximum number of any one type we could take is six. For example, if we take all six cookies as chocolate chip, that's allowed because there are at least six chocolate chip cookies. Similarly for the others. Therefore, there are no restrictions beyond the non-negative integers.

Therefore, the answer is indeed 28.

But let me try another approach to verify. Let's think about enumerating the possibilities. For example, the number of ways to have 0 chocolate chip cookies: then we have to choose 6 cookies from oatmeal and peanut butter. That would be the number of non-negative solutions to o + p = 6, which is C(6 + 2 - 1, 2 - 1) = C(7, 1) = 7.

Similarly, for 1 chocolate chip cookie: then o + p = 5, which is C(5 + 2 - 1, 2 - 1) = C(6, 1) = 6.

For 2 chocolate chip cookies: o + p = 4, which is C(5, 1) = 5.

Continuing this way:

3 chocolate chip: o + p = 3: C(4, 1) = 4

4 chocolate chip: o + p = 2: C(3, 1) = 3

5 chocolate chip: o + p = 1: C(2, 1) = 2

6 chocolate chip: o + p = 0: C(1, 1) = 1

Now, adding all these up: 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28.

Yes, that's the same answer. So that seems to confirm it.

Alternatively, if I think of the formula for combinations with repetition, which is C(n + k - 1, k - 1). Here n is 6, k is 3, so C(6 + 3 - 1, 3 - 1) = C(8, 2) = 28.

So all methods lead to 28. Therefore, I think the answer is 28.

But just to be thorough, let me think if there's another way someone might approach this problem and perhaps get a different answer. For example, if someone thought that since there are three types, maybe they would think of permutations or something else. But no, since the cookies are not distinguishable except by type, it's purely a matter of how many of each type.

Alternatively, someone might think that since there are three types, the number of assortments is 3^6, but that would be if the cookies were distinguishable, which they are not. So that approach is wrong.

Alternatively, someone might think that since there are three types, the number of ways is similar to arranging letters, but again, that's not the case here.

Alternatively, maybe someone might think that since there are at least six of each type, but since we are only choosing six, maybe there's some restriction? But no, because even if you take all six of one type, there are enough cookies. So no restrictions.

Therefore, I think the answer is 28.

Wait, but let me check with another formula. The number of non-negative integer solutions to c + o + p = 6 is equal to the combination with repetition formula. Which is indeed C(6 + 3 - 1, 3 - 1) = C(8, 2) = 28.

Alternatively, if I use generating functions. The generating function for each type is 1 + x + x^2 + x^3 + ... since there are unlimited cookies (but in our case, since we are only choosing 6, even if there were a limited number, but since there are at least 6, it's okay). So the generating function is (1/(1 - x))^3. We need the coefficient of x^6 in this expansion.

The coefficient of x^n in (1/(1 - x))^k is C(n + k - 1, k - 1). Therefore, coefficient of x^6 is C(6 + 3 - 1, 3 - 1) = C(8, 2) = 28.

Same answer.

Therefore, I think I can be confident that the answer is 28.

But just to make sure, let me think of a smaller case. Suppose instead of 6 cookies, we had 2 cookies, and 3 types. How many assortments?

Using the formula, it would be C(2 + 3 - 1, 3 - 1) = C(4, 2) = 6.

Let me list them:

(0,0,2), (0,1,1), (0,2,0), (1,0,1), (1,1,0), (2,0,0). That's 6.

Yes, that's correct. So the formula works here.

Similarly, if we have 3 cookies, the formula would give C(3 + 3 -1, 3 -1) = C(5, 2) = 10.

Listing them would be tedious, but I can trust the formula.

Therefore, going back to the original problem, 6 cookies, 3 types, answer is 28.

Therefore, I think the answer 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 count the number of **non-negative integer solutions** to the equation:

$$
c + o + p = 6
$$

where:
- $ c $ is the number of chocolate chip cookies,
- $ o $ is the number of oatmeal cookies,
- $ p $ is the number of peanut butter cookies.

Each variable represents a non-negative integer (i.e., $ c, o, p \geq 0 $), and the total number of cookies selected is 6.

---

### Step 1: Recognize the Problem Type

This is a classic **combinations with repetition** problem. We are choosing 6 items (cookies) from 3 types (chocolate chip, oatmeal, peanut butter), and the order of selection does not matter. Since cookies of the same type are indistinguishable, we are counting the number of **distinct combinations** of counts of each type that sum to 6.

This is equivalent to finding the number of **non-negative integer solutions** to the equation:

$$
c + o + p = 6
$$

---

### 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 the **stars and bars** formula:

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

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

So the number of solutions is:

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

---

### Step 3: Compute the Value

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

---

### Step 4: Interpret the Result

Each of these 28 combinations corresponds to a unique **assortment** of six cookies, where the counts of each type of cookie are non-negative integers that sum to 6. Since the problem states there are at least six of each type of cookie, we don't need to worry about exceeding the available quantity of any 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 $3\cdot 3 + 3\cdot 6 + 1 = \boxed{28}$.