2021 AMC Fall 10A Problem 21

Every toss sequence is one of $5^{20}$ equally likely outcomes, so $p/q$ is just the ratio of the numbers of favorable outcomes.

For $q$: assign the $20$ distinguishable balls to bins with sizes $4, 4, 4, 4, 4$. The count is the multinomial coefficient
$$N_q = \frac{20!}{(4!)^5}.$$

For $p$: first decide which bin gets $3$ balls ($5$ choices) and which gets $5$ ($4$ remaining choices); the other three get $4$ each. Then assign balls:
$$N_p = 5 \cdot 4 \cdot \frac{20!}{3!\,5!\,(4!)^3}.$$

Divide; the $20!$ cancels:
$$\frac{p}{q} = \frac{N_p}{N_q} = 20 \cdot \frac{(4!)^5}{3!\,5!\,(4!)^3} = 20 \cdot \frac{4!}{3!} \cdot \frac{4!}{5!} = 20 \cdot 4 \cdot \frac{1}{5} = 16.$$

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

When two events live in the same uniform sample space, their probability ratio is a ratio of counts, and multinomial coefficients count "how many ways to hand out labeled objects into bins of given sizes." Almost everything cancels in the ratio; what survives is the number of ways to label which bins are special ($20$) times the small factorial correction $\frac{4!\,4!}{3!\,5!} = \frac45$.

Double counting. Start from any all-fours outcome and move one ball: $20$ choices of ball, $4$ choices of new bin, giving $80$ ways, each producing a $3$-$5$-$4$-$4$-$4$ outcome. Each $3$-$5$-$4$-$4$-$4$ outcome is produced exactly $5$ times (any of the $5$ balls in the full bin could have been the one moved from the light bin). Hence $80 N_q = 5 N_p$, so $N_p / N_q = 16$.

• Forgetting the factor 5 * 4 = 20 for which bin holds 3 balls and which holds 5, which gives the ratio 4/5 and no matching choice, or using 5!/3! = 20 alone.
• Inverting the small factor and computing $20 \cdot \tfrac{5}{4} = 25$, or treating the balls as indistinguishable so that every distribution looks equally likely.