Difference between revisions of "2021 Fall AMC 10A Problems/Problem 21"
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| + | ==Problem== | ||
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>? | 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>? | ||
<math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ | <math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 8 \qquad\textbf{(D)}\ | ||
12 \qquad\textbf{(E)}\ 16</math> | 12 \qquad\textbf{(E)}\ 16</math> | ||
| + | |||
| + | ==Solution 1 (Multinomial Numbers)== | ||
| + | For simplicity purposes, we assume that the balls are indistinguishable and the bins are distinguishable. | ||
| + | |||
| + | Let <math>d</math> be the number of ways to distribute <math>20</math> balls to <math>5</math> bins. We have | ||
| + | <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> | ||
| + | |||
| + | <i><b>Remark</b></i> | ||
| + | |||
| + | By the stars and bars argument, we get <math>d=\binom{20+5-1}{5-1}=\binom{24}{4}.</math> | ||
| + | |||
| + | ~MRENTHUSIASM | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2021 Fall|ab=A|num-b=20|num-a=22}} | ||
| + | {{MAA Notice}} | ||
Revision as of 19:51, 22 November 2021
Problem
Each of the 
balls is tossed independently and at random into one of the 
bins. Let 
be the probability that some bin ends up with 
balls, another with balls, and the other three with 
balls each. Let 
be the probability that every bin ends up with balls. What is 
?
Solution 1 (Multinomial Numbers)
For simplicity purposes, we assume that the balls are indistinguishable and the bins are distinguishable.
Let 
be the number of ways to distribute balls to bins. We have
![\[p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{d} \text{ and } q=\frac{\binom{20}{4,4,4,4,4}}{d}.\]](http://latex.artofproblemsolving.com/a/c/c/acce6de0326df878c34b9393d7b55ff361bf8e0d.png)
Therefore, the answer is ![\[\frac pq=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{\binom{20}{4,4,4,4,4}}=\frac{5\cdot4\cdot\frac{20!}{3!5!4!4!4!}}{\frac{20!}{4!4!4!4!4!}}=\frac{5\cdot4\cdot(4!4!4!4!4!)}{3!5!4!4!4!}=\frac{5\cdot4\cdot4}{5}=\boxed{\textbf{(E)}\ 16}.\]](http://latex.artofproblemsolving.com/8/7/f/87f26edfc03a25e4356549bdf4b3da8758e66cea.png)
Remark
By the stars and bars argument, we get 
~MRENTHUSIASM
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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