2021 AMC 10B Problem 22

Step 1: fix Ang's arrangement.

1. Each person puts their $5$ colors into the $5$ boxes in some order. There are $5 \times 4 \times 3 \times 2 \times 1 = 120$ orders.
2. Whatever Ang does, we can name the boxes by the color Ang put in them: the "red box," the "blue box," and so on. So think of Ang as always putting red in the red box, blue in the blue box, etc.
3. A box gets three blocks of one color exactly when Ben and Jasmin both put that box's color in it. Call such a box a match.
4. Ben and Jasmin together have $120 \times 120 = 14400$ equally likely pairs of orders. We count the pairs with at least one match.

Step 2: count pairs where chosen boxes match.

1. Pick one box, say the red box. Ben must put red there, and his other $4$ colors can go anywhere: $4! = 24$ ways. The same for Jasmin, so $24 \times 24 = 576$ pairs make the red box a match.
2. In the same way, if we pick $k$ particular boxes that must all match, each person has $(5 - k)!$ ways to place the rest, so there are $\big((5-k)!\big)^2$ pairs.

Step 3: add and subtract to fix double counting.

7. Adding $5 \times 576$ counts a pair with two matches twice, a pair with three matches three times, and so on. The standard fix (inclusion–exclusion) is to subtract the "two boxes" count, add back the "three boxes" count, and keep alternating:
$$5 \cdot 576 - 10 \cdot 36 + 10 \cdot 4 - 5 \cdot 1 + 1 \cdot 1 = 2880 - 360 + 40 - 5 + 1 = 2556.$$
8. After this, every pair with at least one match is counted exactly once.

Step 4: simplify.

1. The probability is $\frac{2556}{14400}$. Both numbers divide by $36$: $\frac{71}{400}$. Since $71$ is prime and does not divide $400$, this is in lowest terms.
2. So $m + n = 71 + 400 = 471$.

The answer is $\boxed{\textbf{(D)}\ 471}$.

Because everyone arranges at random, one person's arrangement can be treated as fixed without changing any chances. The events "this box matches" overlap, so simply adding their counts overcounts. Adding, then subtracting the overlaps, then adding back the triple overlaps, and so on, corrects this. A quick check: adding five chances of $\frac{1}{25}$ gives $0.2$, an overcount, and the true answer $\frac{71}{400} = 0.1775$ is a little smaller.

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 $\boxed{\textbf{(D)}\ 471}$.