2003 AMC 10A Problem 21

Two oatmeal cookies are the same as any other two oatmeal cookies, so an assortment is described completely by three counts: how many chocolate chip, how many oatmeal, how many peanut butter. These are whole numbers (zero allowed) that add to $6$. There are at least six of each kind, so every such split is possible.

The cleanest way to list them is by the number of chocolate chip cookies.

1. 0 chocolate chip. The other $6$ are oatmeal and peanut butter: oatmeal can be $0, 1, 2, 3, 4, 5$ or $6$, and peanut butter is the rest. That is $7$ assortments.
2. 1 chocolate chip. The other $5$ are split between two kinds: oatmeal can be $0$ to $5$. That is $6$ assortments.
3. 2 chocolate chip. The other $4$ split in $5$ ways.
4. 3 chocolate chip: $4$ ways. 4 chocolate chip: $3$ ways. 5 chocolate chip: $2$ ways. 6 chocolate chip: $1$ way.

Total:
$$7 + 6 + 5 + 4 + 3 + 2 + 1 = 28.$$

The answer is $\boxed{\textbf{(D)}\ 28}$.

When the items of each kind are identical, only the counts matter, so the question becomes "how many ways can a few whole numbers add up to a total?" Fixing one count at a time turns it into a short, orderly list. The phrase "different assortments" is the signal that order does not matter and the cookies of one kind are interchangeable.

Stars and bars. Draw the $6$ cookies as $6$ stars in a row and put $2$ bars among them to split them into three groups: chocolate chip, then oatmeal, then peanut butter. For example
$$\star\star\,|\,\star\star\star\,|\,\star$$
means $2$, $3$, $1$, and $\star\star\star\star\star\star\,|\,|$ means all $6$ are chocolate chip. Each row of $8$ symbols ($6$ stars, $2$ bars) matches one assortment, and each assortment matches one row. Choosing which $2$ of the $8$ spots hold the bars gives $\frac{8 \times 7}{2} = 28$ rows.

Cookies of the same kind are interchangeable, so an assortment is determined by the counts $(a, b, c)$ of chocolate chip, oatmeal and peanut butter cookies, which are nonnegative integers with $a + b + c = 6$. The "at least six of each" clause guarantees every such triple is actually available.

Represent a triple as a row of $6$ stars split by $2$ bars: for example $\star\star\,|\,\star\star\star\star\,|$ means $(2, 4, 0)$. Every arrangement of $8$ symbols with $2$ bars corresponds to exactly one assortment, so the count is
$$\binom{8}{2} = 28.$$

The answer is $\boxed{\textbf{(D)}\ 28}$.