2021 AMC 10B Spring Problem 22

Question

Ang, Ben, and Jasmin each have

5

blocks, colored red, blue, yellow, white, and green; and there are

5

empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives

3

blocks all of the same color is

\frac{m}{n},

where

m

and

n

are relatively prime positive integers. What is

m + n ?

47

94

227

471

542

Answer: D

Concepts:inclusion-exclusion permutations basic probability

Difficulty rating: 2150

Small hint:

Fix Ang’s placement and count Ben’s and Jasmin’s permutations

Big hint:

Use inclusion-exclusion on the boxes where both other placements match Ang’s color

Video solution:

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Written solution:

Fix Ang’s placement and label each box by the color Ang put in it. Ben and Jasmin each choose a permutation of the five colors, so there are

(5!)^2

equally likely pairs of placements.

For a specified set of

k

boxes to receive three blocks of the same color, both Ben and Jasmin must match Ang in those

k

boxes. This can happen in

((5-k)!)^2

ways. By inclusion-exclusion, the number of successful placement pairs is

\begin{aligned} &\binom51(4!)^2-\binom52(3!)^2 \\ &\quad {}+\binom53(2!)^2-\binom54(1!)^2 \\ &\quad {}+\binom55(0!)^2. \end{aligned}

This equals

2880-360+40-5+1=2556.

Therefore the probability is

\frac{2556}{(5!)^2}=\frac{2556}{14400}=\frac{71}{400}.

Thus

m+n=71+400=471.

Thus, the answer is D .