2001 AMC 10 Problem 19

Only the counts matter: a selection is a triple of nonnegative integers $(g, c, p)$ with $g + c + p = 4$.

Encode each triple as a row of $4$ identical stars split into three groups by $2$ bars; for example $\ast\ast\,|\,\ast\,|\,\ast$ means two glazed, one chocolate, one powdered, and $|\,|\,\ast\ast\ast\ast$ means four powdered. Every arrangement of $4$ stars and $2$ bars corresponds to exactly one selection, and there are
$$\binom{6}{2} = 15$$
such arrangements (choose which $2$ of the $6$ slots are bars).

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

Selecting $n$ items from $k$ types with unlimited repetition is counted by $\binom{n + k - 1}{k - 1}$; the stars-and-bars picture is the reason the formula works and is easy to rebuild under pressure. Recognize the setup by the phrases "ample supply" and "how many different selections," which signal unordered, repetition allowed.

Count by how the four donuts split among types: $4{+}0{+}0$ ($3$ ways to pick the type), $3{+}1{+}0$ ($3 \cdot 2 = 6$ ways), $2{+}2{+}0$ ($3$ ways), $2{+}1{+}1$ ($3$ ways). Total $3 + 6 + 3 + 3 = 15$.

• Treating the donuts as distinguishable (3^4 orderings) or forgetting selections like four of one kind.
• Using $\binom{6}{3}$ or $\binom{7}{3}$ from a misremembered formula, or requiring at least one of each type (which gives only $3$).