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2001 AMC 10 Problems/Problem 19

Revision as of 10:17, 8 November 2021 by Dairyqueenxd (talk | contribs) (→‎Solution 2)

Problem

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?

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

Solution

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 stars and bars (balls and urns) 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 $\binom{4}{3}=6.$ Thus, our answer is $3 \cdot 3+6 = \boxed{\textbf{(D) }15}$.

Solution by IronicNinja

Edit by virjoy2001 (Reason LaTeX mistake)

See Also

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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