Difference between revisions of "2001 AMC 10 Problems/Problem 19"

Latest revision as of 16:35, 7 June 2024

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 1

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

Similar to Solution 1, we can generalize the problem by using the stars and bars formula, yielding an answer of $\dbinom{4+3-1}{3-1}=\dbinom{6}{2}=\boxed{\textbf{(D)}\ 15}$ .

Solution 3

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 $\frac{3!}{2!1!}$ 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 & Aopsthedude

@@ Line 5: / Line 5: @@
 <math> \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 18 </math>
-== Solution==
+== Solution 1 ==
 Let's use [[stars and bars]].
-Let the donuts be represented by <math> O </math>s. We wish to find all possible combinations of glazed, chocolate, and powdered donuts that give us <math> 4 </math> in all. The four donuts we want can be represented as <math> OOOO </math>. 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, <math> O|OO|O </math> 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 <math> \binom{6}{2}=15 </math> ways. Our answer is hence <math> \boxed{\textbf{(D)}\ 15} </math>. Notice that this can be generalized to get the balls and urn (stars and bars) identity.
+Let the donuts be represented by <math> O </math>s. We wish to find all possible combinations of glazed, chocolate, and powdered donuts that give us <math> 4 </math> in all. The four donuts we want can be represented as <math> OOOO </math>. 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, <math> O|OO|O </math> 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 <math> \binom{6}{2}=15 </math> ways. Our answer is hence <math> \boxed{\textbf{(D)}\ 15} </math>. Notice that this can be generalized to get the stars and bars (balls and urns) identity.
 ==Solution 2==
+Similar to Solution 1, we can generalize the problem by using the stars and bars formula, yielding an answer of <math>\dbinom{4+3-1}{3-1}=\dbinom{6}{2}=\boxed{\textbf{(D)}\ 15}</math>.
+==Solution 3==
 Simple casework works here as well:
@@ Line 19: / Line 22: @@
 <cmath>2:1:1</cmath>
-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 <math>\frac{4!}{3!2!}</math> 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 <math>3*3+6 = \boxed{15}</math>
+In three of these cases we see that there are two of the same ratios (so like two boxes would have <math>0</math>), and so if we swapped those two donuts, we would have the same case. Thus we get <math>\frac{3!}{2!1!}</math> for those <math>3</math> (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 <math>\binom{4}{3}=6.</math> Thus, our answer is <math>3 \cdot 3+6 = \boxed{\textbf{(D) }15}</math>.
 Solution by IronicNinja
+Edit by virjoy2001 & Aopsthedude
-==Solution 3==
-Since there are three choices to pick from, the value of <math>a</math> in the expression <math>{a \choose b}</math> has to be a multiple of <math>3</math>. <math>{3 \choose 4}</math> would make no sense, because that would mean: "How many ways are there to choose 4 donuts out of 3?". Next we could try <math>{6 \choose 4}</math>, which is equal to <math>15</math> ways. This works, but now let <math>a</math> be <math>9</math>. So if there were <math>9</math> donuts, (which is possible since it's a multiple of 3), the number of ways we could choose <math>4</math> donuts would be <math>{9 \choose 4} = 126</math>. Anything greater than <math>18</math> will be incorrect, since <math>18</math> is the largest number in the answer choices. Thus, our answer is <math>\boxed {15}</math>.
 == 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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