2003 AMC 10A Problem 21

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 $\boxed{\textbf{(D)}\ 28}$.

Choosing $n$ items from $k$ types with repetition allowed is $\binom{n+k-1}{k-1}$. The bijection with star-bar strings converts an unordered multiset problem into counting positions for bars. Recognize the setup by the phrase "how many different assortments" together with an unlimited supply of each type.

Casework by pattern of counts: one kind only ($3$ ways); two kinds with splits $5+1, 4+2, 3+3$ ($3 \cdot 5 = 15$ ordered ways across the $3$ pairs of kinds); three kinds with multisets $\{4,1,1\}, \{3,2,1\}, \{2,2,2\}$ giving $3 + 6 + 1 = 10$. Total $3 + 15 + 10 = 28$.

• Treating the six cookies as distinguishable and answering 3^6 = 729, choice (E).
• Using $\binom{7}{2} = 21$ by forgetting that two bars may sit together (a kind may be chosen zero times).