2001 AMC 10 Problem 19

1. A "selection" is just how many of each type Pat buys. Picking glazed then chocolate is the same as chocolate then glazed. So we count the ways to split $4$ donuts into glazed, chocolate and powdered amounts, where an amount can be $0$.
2. Organize by the shape of the split, ignoring which type gets which amount:
- $4, 0, 0$: all four are one type. The type can be any of $3$, so $3$ ways.
- $3, 1, 0$: choose the type that gets $3$ ($3$ choices), then the type that gets $1$ ($2$ choices left). That is $3 \times 2 = 6$ ways.
- $2, 2, 0$: choose the type that gets none ($3$ choices). That is $3$ ways.
- $2, 1, 1$: choose the type that gets $2$ ($3$ choices). That is $3$ ways.
3. These are all the ways to write $4$ as a sum of three amounts, so the total is
$$3 + 6 + 3 + 3 = 15.$$

The answer is $\boxed{\textbf{(D)}\ 15}$.

"Ample supply" means you can take as many of one type as you like, and "selections" means order does not matter. So the question is really how many ways three amounts can add to $4$. Sorting by the shape of the split (like $3, 1, 0$) keeps the list complete without repeats.

Draw the $4$ donuts as dots and add $2$ dividers to separate the types: glazed before the first divider, chocolate between them, powdered after the second. For example, $\bullet\bullet \mid \bullet \mid \bullet$ is $2$ glazed, $1$ chocolate, $1$ powdered, and $\mid \mid \bullet\bullet\bullet\bullet$ is $4$ powdered. Every selection is one row of $6$ symbols ($4$ dots and $2$ dividers), and every such row is one selection. Choosing which $2$ of the $6$ spots hold dividers gives $\frac{6 \times 5}{2} = 15$ rows.