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| − | ==Problem==
| + | #REDIRECT [[2021_Fall_AMC_12A_Problems/Problem_18]] |
| − | Each of the <math>20</math> balls is tossed independently and at random into one of the <math>5</math> bins. Let <math>p</math> be the probability that some bin ends up with <math>3</math> balls, another with <math>5</math> balls, and the other three with <math>4</math> balls each. Let <math>q</math> be the probability that every bin ends up with <math>4</math> balls. What is <math>\frac{p}{q}</math>?
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| − | <math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\
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| − | 12 \qquad\textbf{(E)}\ 16</math>
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| − | ==Solution 1 (Multinomial Numbers)==
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| − | For simplicity purposes, we assume that the balls are indistinguishable and the bins are distinguishable.
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| − | Let <math>d</math> be the number of ways to distribute <math>20</math> balls to <math>5</math> bins. We have
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| − | <cmath>p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{d} \text{ and } q=\frac{\binom{20}{4,4,4,4,4}}{d}.</cmath> Therefore, the answer is <cmath>\frac pq=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{\binom{20}{4,4,4,4,4}}=\frac{5\cdot4\cdot\frac{20!}{3!5!4!4!4!}}{\frac{20!}{4!4!4!4!4!}}=\frac{5\cdot4\cdot(4!4!4!4!4!)}{3!5!4!4!4!}=\frac{5\cdot4\cdot4}{5}=\boxed{\textbf{(E)}\ 16}.</cmath>
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| − | <u><b>Remark</b></u>
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| − | By the stars and bars argument, we get <math>d=\binom{20+5-1}{5-1}=\binom{24}{4}.</math>
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| − | ~MRENTHUSIASM
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| − | ==Solution 2 (Simple) ==
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| − | ==See Also==
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| − | {{AMC10 box|year=2021 Fall|ab=A|num-b=20|num-a=22}}
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| − | {{MAA Notice}}
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