2021 Fall AMC 10A Problems/Problem 21
Contents
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 Coefficients)
For simplicity purposes, we assume that the balls are indistinguishable and the bins are distinguishable.
Let 
be the number of ways to distribute balls into 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
Solution 2 (Simple)
Since both of the boxes will have boxes with balls in them, we can leave those out. There are 
= ways to choose where to place the and the . After that, there are 
ways to put the and balls being put into the boxes. For the 
case, after we canceled the 
out, we have 
= 
ways to put the balls inside the boxes. Therefore, we have 
which is equal to 
= 
~Arcticturn
Video Solution by Punxsutawney Phil
https://YouTube.com/watch?v=bvd2VjMxiZ4
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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