2021 AMC Fall 10A Problem 21

1. Same bottom. Each ball can land in $5$ bins, so there are $5^{20}$ equally likely ways for the balls to land. Both $p$ and $q$ are "good ways divided by $5^{20}$". So
$$\frac{p}{q} = \frac{\text{number of ways for the 3-5-4-4-4 pattern}}{\text{number of ways for the 4-4-4-4-4 pattern}}.$$
2. A counting tool. How many ways can labeled balls be split into bins with given sizes? Small example: $4$ balls into two bins, sizes $1$ and $3$. You pick which ball goes alone: $4$ ways. The formula $\frac{4!}{1!\,3!} = \frac{24}{6} = 4$ agrees. In general, the number of ways is
$$\frac{(\text{total})!}{(\text{size}_1)!\,(\text{size}_2)!\cdots}.$$
(Line up all the balls, $20!$ orders, and put the first few in bin 1, the next few in bin 2, and so on. Reordering balls inside a bin does not change anything, so divide by those factorials.)
3. Count for $q$. All five bins get $4$ balls:
$$\frac{20!}{4!\,4!\,4!\,4!\,4!}.$$
4. Count for $p$. First choose which bin gets $3$ balls ($5$ choices), then which gets $5$ balls ($4$ choices left). That is $5 \times 4 = 20$ ways. The other three bins get $4$ each. Then split the balls:
$$20 \times \frac{20!}{3!\,5!\,4!\,4!\,4!}.$$
5. Divide. The $20!$ cancels, and three of the $4!$s cancel:
$$\frac{p}{q} = 20 \times \frac{4!\,4!}{3!\,5!} = 20 \times \frac{4!}{3!} \times \frac{4!}{5!} = 20 \times 4 \times \frac{1}{5} = 16.$$

The answer is $\boxed{\textbf{(E)}\ 16}$.

When two events come from the same set of equally likely outcomes, the ratio of their probabilities is just the ratio of their counts, so the huge $5^{20}$ never needs computing. In the counts, almost everything cancels. What is left is the number of ways to pick the two special bins ($20$) and a small factor $\frac45$ from changing one bin from $4$ to $3$ and another from $4$ to $5$.

Count by moving one ball. Start from any all-fours outcome. Move one ball ($20$ choices) to a different bin ($4$ choices). That makes $80$ moves, and each one gives a 3-5-4-4-4 outcome. Now go backwards: a given 3-5-4-4-4 outcome can be reached this way in exactly $5$ ways, since any of the $5$ balls in the full bin could have been the one that moved in from the bin with $3$. So
$$80 \times (\text{ways for } q) = 5 \times (\text{ways for } p),$$
and $\frac{p}{q} = \frac{80}{5} = 16$.