<Question>
Pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each of these three kinds of cookies on the tray. How many different assortments of six cookies can be selected? (Note that cookies of the same type are not distinguishable.)
</Question>

<think>
<Plan_1>
Okay, so Pat needs to select six cookies from a tray that has chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each type, so we don't have to worry about running out. The question is asking how many different assortments Pat can select. Hmm, alright. Let me think about how to approach this.

First, since the cookies of the same type are not distinguishable, the problem is about combinations where the order doesn't matter, and we have multiple types. This sounds like a classic combinatorics problem, specifically a problem of combinations with repetition, also known as "stars and bars".

<prompt_1.1> Identify the type of combinatorics problem and set up the initial equation using the stars and bars method. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Let me recall the formula for combinations with repetition. If there are n types of items and we want to choose k items, the number of ways is C(n + k - 1, k). Alternatively, sometimes it's written as C(n + k - 1, n - 1), because the formula is symmetric. Either way, the formula is used when we can have multiple items of the same type and the order doesn't matter.

In this problem, the types are chocolate chip, oatmeal, and peanut butter. So n = 3. The number of cookies to choose is 6, so k = 6. Therefore, plugging into the formula, we get C(3 + 6 - 1, 6) = C(8, 6). Alternatively, C(8, 3) because C(8,6) = C(8,2), but wait, let me check that.

Wait, the formula is C(n + k - 1, k) which is the same as C(n + k - 1, n - 1). So here, n is 3, k is 6. So it's C(3 + 6 - 1, 6) = C(8,6) or equivalently C(8, 3 - 1) = C(8,2). Wait, no, hold on. Let me clarify. The formula is C(n + k - 1, k) or C(n + k - 1, n - 1). Both are equal because C(a, b) = C(a, a - b). So C(8,6) = C(8,2). Let me compute both to check.

C(8,6) = 8! / (6! 2! ) = (8 × 7) / (2 × 1) = 28.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
C(8,2) = 8! / (2! 6! ) = same thing, 28. So either way, the answer is 28.
So there are 28 different ways.

Wait, but let me make sure I'm not missing anything here. The problem states that there are at least six of each type, so we don't have to worry about the case where we might select more than the available quantity. Since each type has at least six cookies, selecting six cookies in total can't exceed the available supply for any type. Therefore, the standard stars and bars formula applies here without any restrictions.

So the answer is 28.

But just to be thorough, let me think through an example.

<prompt_2.1> Use a smaller example to verify the solution and ensure the reasoning is correct. </prompt_2.1>

Alternatively, let's think of the problem as 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 oatmeal, and p is peanut butter.

<prompt_2.2> Use a different perspective to confirm the solution and reasoning. </prompt_2.2>

Another way to think about it is by using the stars and bars model. Imagine the 6 cookies as stars: ***

<prompt_2.3> Provide a visual representation using stars and bars to confirm the solution and reasoning. </prompt_2.3>

But let me make sure there are no alternative interpretations of the problem that could change the answer. For example, does the problem consider different types distinguishable?

<prompt_2.4> Consider alternative interpretations and confirm the solution's validity. </prompt_2.4>
</Plan_2>

<Execution_2>
<execution_2.1>
Suppose instead there were 3 cookies and 2 types. Then the number of assortments would be C(2 + 3 -1, 3) = C(4,3) = 4. Let's list them: 0 and 3, 1 and 2, 2 and 1, 3 and 0. That's four, which matches. So the formula works here. Similarly, in our original problem, 3 types and 6 cookies, gives 28.
</execution_2.1>

<execution_2.2>
Each variable is greater than or equal to zero since you can select zero cookies of a type. 

The stars and bars theorem tells us that the number of solutions is C(n + k - 1, k) where n is the number of variables (types) and k is the number of items (cookies). Wait, actually, hold on. Let me check the formula again. 

Wait, in the stars and bars theorem, the formula for the number of non-negative integer solutions to x₁ + x₂ + ... + xₙ = k is C(n + k - 1, k). So here, n is 3 (the three types), and k is 6 (the six cookies). So it's C(3 + 6 - 1, 6) = C(8,6) = 28. 

Alternatively, sometimes the formula is presented as C(n + k - 1, n - 1). So in this case, that would be C(3 + 6 - 1, 3 - 1) = C(8,2) = 28. Same result. 

Therefore, the number of different assortments is 28.
</execution_2.2>

<execution_2.3>
We need to separate them into three groups (chocolate, oatmeal, peanut butter) using two bars |. For example, **|*|** would mean 2 chocolate, 1 oatmeal, and 2 peanut butter. The total number of symbols is 6 stars + 2 bars = 8 symbols. The number of ways to arrange these is C(8,2) because we choose positions for the bars (or equivalently the stars). Which is 28.
</execution_2.3>

<execution_2.4>
But the problem says "cookies of the same type are not distinguishable," so when we select, say, 2 chocolate chip and 3 oatmeal, that's considered one unique assortment, not multiple because the cookies are indistinct. So yes, combinations with repetition.

Alternatively, if the cookies were distinguishable, the answer would be different, but they are not. So the stars and bars approach is correct here.

Another check: if there were only one type of cookie, the number of ways would be 1 (just 6 of that one type). If there were two types, the formula would be C(2 + 6 -1, 6) = C(7,6) = 7. Let's verify that. For two types, the number of ways to choose 6 is the number of non-negative integer solutions to a + b = 6, which are (0,6), (1,5), ...., (6,0), total 7. Correct. So the formula works for two types.

Therefore, for three types, the formula gives 28, which should be correct. So I think the answer is 28.
</execution_2.4>
</Execution_2>

<Final_answer>
Therefore, there are $\boxed{28}$ different assortments of six cookies that Pat can select.
</Final_answer>
</think>