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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 Coefficients)==
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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 into <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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| − | Since both of the boxes will have <math>3</math> boxes with <math>4</math> balls in them, we can leave those out. There are <math>\binom {6}{3}</math> = <math>20</math> ways to choose where to place the <math>3</math> and the <math>5</math>. After that, there are <math>\binom {8}{3} = 56</math> ways to put the <math>3</math> and <math>5</math> balls being put into the boxes. For the <math>4,4,4,4,4</math> case, after we canceled the <math>4,4,4</math> out, we have <math>\binom {8}{4}</math> = <math>70</math> ways to put the <math>4</math> balls inside the boxes. Therefore, we have <math>\frac {56\cdot 20}{70}</math> which is equal to <math>8 \cdot 2</math> = <math>\boxed {(E)16}</math>
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| − | ~Arcticturn
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| − | ==Solution 3 (30-second set-theoretic solution) ==
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| − | Construct the set <math>A</math> consisting of all possible <math>3-5-4-4-4</math> bin configurations, and construct set <math>B</math> consisting of all possible <math>4-4-4-4-4</math> configurations. If we let <math>N</math> be the total number of configurations possible, it's clear we want to solve for <math>\frac{p}{q} = \frac{\frac{|A|}{N}}{\frac{|B|}{N}} = \frac{|A|}{|B|}</math>.
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| − | Consider drawing an edge between an element in <math>A</math> and an element in <math>B</math> if it is possible to reach one configuration from the other by moving a single ball (note this process is reversible). Let us consider the total number of edges drawn.
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| − | From any element in <math>A</math>, we may take one of the <math>5</math> balls in the 5-bin and move it to the 3-bin to get a valid element in <math>B</math>. This implies the number of edges is <math>5|A|</math>.
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| − | On the other hand for any element in <math>B</math>, we may choose one of the <math>20</math> balls and move it to one of the other <math>4</math> bins to get a valid element in <math>A</math>. This implies the number of edges is <math>80|B|</math>.
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| − | Since they must be equal, then <math>5|A| = 80|B| \rightarrow \frac{|A|}{|B|} = \frac{80}{5} = \boxed {(E)16}</math>
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| − | ==Video Solution by Mathematical Dexterity==
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| − | https://www.youtube.com/watch?v=Lu6eSvY6RHE
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| − | ==Video Solution by Punxsutawney Phil==
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| − | https://YouTube.com/watch?v=bvd2VjMxiZ4
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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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