2021 AMC 10B Problem 22

Each person's placement is a permutation of the five colors. Relabel the colors by Ang's placement, so Ang puts color $i$ in box $i$. Ben's and Jasmin's placements are then independent uniformly random permutations $\sigma$ and $\tau$, and box $i$ is monochromatic exactly when $\sigma(i) = i$ and $\tau(i) = i$: box $i$ is a common fixed point.

Count ordered pairs $(\sigma, \tau)$ out of $120^2 = 14400$ that share at least one fixed point. For a chosen set of $k$ boxes, the number of pairs in which both permutations fix all $k$ of them is $\big((5-k)!\big)^2$. Inclusion–exclusion over $k = 1, \dots, 5$:
$$\binom51 (4!)^2 - \binom52 (3!)^2 + \binom53 (2!)^2 - \binom54 (1!)^2 + \binom55 (0!)^2$$
$$= 5 \cdot 576 - 10 \cdot 36 + 10 \cdot 4 - 5 \cdot 1 + 1 = 2880 - 360 + 40 - 5 + 1 = 2556.$$

The probability is $\frac{2556}{14400}$. Dividing by $36$: $\frac{71}{400}$, and $71$ is prime, so this is lowest terms.

$m + n = 71 + 400 = 471$. The answer is $\boxed{\textbf{(D)}\ 471}$.

Symmetry lets you fix one player's arrangement for free, turning "three blocks match" into "two permutations agree with the identity at some position." Events like "box $i$ matches" overlap heavily, so inclusion–exclusion is the natural tool, and fixing $k$ specified points of a permutation leaves $(5-k)!$ freedom. Sanity check: the union bound gives at most $5 \cdot \frac{1}{25} = 0.2$, and $\frac{71}{400} = 0.1775$ sits just below it.

• Adding the five box probabilities (5 · 1/25) without subtracting overlaps, or dropping the alternating signs in inclusion–exclusion.
• Using $(5-k)!$ instead of $\big((5-k)!\big)^2$, forgetting that both Ben and Jasmin must fix the chosen boxes.