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

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<math> \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 18 </math>
 
<math> \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 18 </math>
  
== Solution ==
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== Solution==
  
Let the donuts be represented by <math> O </math>s. We wish to find all 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 identity.
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Let's use [[stars and bars]].
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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.
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==Solution 2==
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Simple casework works here as well:
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Set up the following ratios:
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<cmath>4:0:0</cmath>
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<cmath>3:1:0</cmath>
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<cmath>2:2:0</cmath>
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<cmath>2:1:1</cmath>
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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{4!}{3!2!}</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{15}</math>.
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Solution by IronicNinja
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Edit by virjoy2001 (Reason LaTeX mistake)
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2001|num-b=18|num-a=20}}
 
{{AMC10 box|year=2001|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 15:09, 30 May 2021

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 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 $\binom{4}{3}=6.$ Thus, our answer is $3 \cdot 3+6 = \boxed{15}$.

Solution by IronicNinja

Edit by virjoy2001 (Reason LaTeX mistake)

See Also

2001 AMC 10 (ProblemsAnswer KeyResources)
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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