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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 21"

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==Solution 2 (Simple) ==
 
==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
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I'm still working on this solution - PLEASE DO NOT EDIT
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~Arcticturn
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=20|num-a=22}}
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:01, 22 November 2021

Problem

Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$?

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 16$

Solution 1 (Multinomial Numbers)

For simplicity purposes, we assume that the balls are indistinguishable and the bins are distinguishable.

Let $d$ be the number of ways to distribute $20$ balls into $5$ bins. We have \[p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{d} \text{ and } q=\frac{\binom{20}{4,4,4,4,4}}{d}.\] Therefore, the answer is \[\frac pq=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{\binom{20}{4,4,4,4,4}}=\frac{5\cdot4\cdot\frac{20!}{3!5!4!4!4!}}{\frac{20!}{4!4!4!4!4!}}=\frac{5\cdot4\cdot(4!4!4!4!4!)}{3!5!4!4!4!}=\frac{5\cdot4\cdot4}{5}=\boxed{\textbf{(E)}\ 16}.\]

Remark

By the stars and bars argument, we get $d=\binom{20+5-1}{5-1}=\binom{24}{4}.$

~MRENTHUSIASM

Solution 2 (Simple)

Since both of the boxes will have $3$ boxes with $4$ balls in them, we can leave those out. There are $\binom {6}{3}$ = $20$ ways to choose where to place the $3$ and the $5$. After that, there are $\binom {8}{3} = 56$ ways to put the $3$ and $5$ balls being put into the boxes. For the $4,4,4,4,4$ case, after we canceled the $4,4,4$ out, we have $\binom {8}{4}$ = $70$ ways to put the $4$ balls inside the boxes. Therefore, we have

I'm still working on this solution - PLEASE DO NOT EDIT

~Arcticturn

See Also

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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