2021 Fall AMC 10A Problem 21

All $5^{20}$ assignments of the distinguishable balls to the labeled bins are equally likely. For $q,$ the number of assignments is $$\frac{20!}{(4!)^5}.$$ For $p,$ choose the bin with $3$ balls and the bin with $5$ balls in $5\cdot4$ ways. The number of assignments is then $$5\cdot4\cdot\frac{20!}{3!5!(4!)^3}.$$ The common probability denominator cancels, so $$\frac pq=20\cdot\frac{(4!)^2}{3!5!}=20\cdot\frac45=16.$$ Thus, E is the correct answer.