# 2021 Fall AMC 12A Problems/Problem 18

The following problem is from both the 2021 Fall AMC 10A #21 and 2021 Fall AMC 12A #18, so both problems redirect to this page.

## 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 Coefficients)

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

Recall that there are $5^{20}$ ways to distribute $20$ balls into $5$ bins. We have $$p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{5^{20}} \text{ and } q=\frac{\binom{20}{4,4,4,4,4}}{5^{20}}.$$ Therefore, the answer is $$\frac pq=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{\binom{20}{4,4,4,4,4}}=\frac{5\cdot4\cdot\frac{20!}{3!\cdot5!\cdot4!\cdot4!\cdot4!}}{\frac{20!}{4!\cdot4!\cdot4!\cdot4!\cdot4!}}=\frac{5\cdot4\cdot(4!\cdot4!\cdot4!\cdot4!\cdot4!)}{3!\cdot5!\cdot4!\cdot4!\cdot4!}=\frac{5\cdot4\cdot4}{5}=\boxed{\textbf{(E)}\ 16}.$$ ~MRENTHUSIASM ~Jesshuang

## Solution 2

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

Let $q=\frac{x}{a},$ where $a$ is the total number of combinations and $x$ is the number of cases where every bin ends up with $4$ balls.

We can take $1$ ball from one bin and place it in another bin so that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Note that one configuration of $4{-}4{-}4{-}4{-}4$ corresponds to $5\cdot4\cdot4=80$ configurations of $3{-}5{-}4{-}4{-}4.$ On the other hand, one configuration of $3{-}5{-}4{-}4{-}4$ corresponds to $5$ configurations of $4{-}4{-}4{-}4{-}4.$

Therefore, we have $$p = \frac{80}{5}\cdot\frac{x}{a} = 16\cdot\frac{x}{a},$$ from which $\frac{p}{q} = \boxed{\textbf{(E)}\ 16}.$

~Hoju

## Solution 3 (Binomial Coefficients)

Since both of the cases will have $3$ bins with $4$ balls in them, we can leave those out. There are $2 \cdot \binom {5}{2} = 20$ 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 bins. 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 bins. Therefore, we have $\frac {56\cdot 20}{70}$ which is equal to $8 \cdot 2 = \boxed{\textbf{(E)}\ 16}$.

~Arcticturn

## Solution 4 (Set Theory / Graph Theory)

Construct the set $A$ consisting of all possible $3{-}5{-}4{-}4{-}4$ bin configurations, and construct set $B$ consisting of all possible $4{-}4{-}4{-}4{-}4$ configurations. If we let $N$ be the total number of configurations possible, it's clear we want to solve for $\frac{p}{q} = \frac{\frac{|A|}{N}}{\frac{|B|}{N}} = \frac{|A|}{|B|}$.

Consider drawing an edge between an element in $A$ and an element in $B$ 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.

For any element in $A$, we may choose one of the $5$ balls in the $5$-bin and move it to the $3$-bin to get a valid element in $B$. This implies the number of edges is $5|A|$.

On the other hand, for any element in $B$, we may choose one of the $20$ balls and move it to one of the other $4$-bins to get a valid element in $A$. This implies the number of edges is $80|B|$.

We equate the expressions to get $5|A| = 80|B|$, from which $\frac{|A|}{|B|} = \frac{80}{5} = \boxed{\textbf{(E)}\ 16}$.

~ pi_is_3.14

~IceMatrix