2001 AMC 10 Problems/Problem 19

Revision as of 11:22, 28 July 2020 by Sakshamsethi (talk | contribs) (Solution 3)


Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?

$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 18$


Let's use stars and bars. Let the donuts be represented by $O$s. We wish to find all possible combinations of glazed, chocolate, and powdered donuts that give us $4$ in all. The four donuts we want can be represented as $OOOO$. Notice that we can add two "dividers" to divide the group of donuts into three different kinds; the first will be glazed, second will be chocolate, and the third will be powdered. For example, $O|OO|O$ represents one glazed, two chocolate, and one powdered. We have six objects in all, and we wish to turn two into dividers, which can be done in $\binom{6}{2}=15$ ways. Our answer is hence $\boxed{\textbf{(D)}\ 15}$. Notice that this can be generalized to get the balls and urn (stars and bars) identity.

Solution 2

Simple casework works here as well: Set up the following ratios: \[4:0:0\] \[3:1:0\] \[2:2:0\] \[2:1:1\]

In three of these cases we see that there are two of the same ratios (so like two boxes would have 0), and so if we swapped those two donuts, we would have the same case. Thus we get $\frac{4!}{3!2!}$ for those 3 (You can also set it up and logically symmetry applies). For the other case where each ratio of donuts is different, we get the normal 4C3 = 6. Thus, our answer is $3*3+6 = \boxed{15}$

Solution by IronicNinja

See Also

2001 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 10 Problems and Solutions

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