2021 AMC 12A Fall Problem 18

Question

Each of

20

balls is tossed independently and at random into one of

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

\dfrac{p}{q}?

1

4

8

12

16

Both probabilities share the same

5^{20}

denominator, so

\dfrac{p}{q}

is a ratio of counts

q

uses

\dfrac{20!}{(4!)^5};

p

uses

5\cdot 4

choices of the special bins times

\dfrac{20!}{3!\,5!\,(4!)^3}

Both probabilities divide by

5^{20},

so

\dfrac{p}{q}

is a ratio of arrangement counts.

For

q,

all bins have

4:

\dfrac{20!}{(4!)^5}.

For

p,

choose which bin has

3

and which has

5

in

5\cdot 4 = 20

ways, times

\dfrac{20!}{3!\,5!\,(4!)^3}.

Therefore

\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.

Problem 18 in other years