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2021 Fall AMC 12A Problem 18

Problem 18 of 25IntermediateCounting & Probability

Each of 2020 balls is tossed independently and at random into one of 55 bins. Let pp be the probability that some bin ends up with 33 balls, another with 55 balls, and the other three with 44 balls each. Let qq be the probability that every bin ends up with 44 balls. What is pq?\dfrac{p}{q}?

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Solution

Both probabilities divide by 520,5^{20}, so pq\dfrac{p}{q} is a ratio of arrangement counts. For q,q, all bins have 4:4: 20!(4!)5.\dfrac{20!}{(4!)^5}. For p,p, choose which bin has 33 and which has 55 in 54=205\cdot 4 = 20 ways, times 20!3!5!(4!)3.\dfrac{20!}{3!\,5!\,(4!)^3}. Therefore pq=20(4!)53!5!(4!)3=20(4!)23!5!=20576720=16. \begin{aligned} \frac{p}{q} &= 20 \cdot \frac{(4!)^5}{3!\,5!\,(4!)^3} \\ &= 20 \cdot \frac{(4!)^2}{3!\,5!} \\ &= 20 \cdot \frac{576}{720} \\ &= 16. \end{aligned} Thus, the correct answer is E.

More practice

Concepts: basic probability · multiset permutations

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.