22. Person A, Person B, and Person C each have five building blocks, colored red, blue, yellow, white, and green, and there are five empty boxes. Now, each person randomly and independently places one of their blocks into each box. If the probability that at least one box contains three blocks of the same color is $\frac{m}{n}\left(m, n \in \mathbf{Z}_{+},(m, n)=1\right)$, then $m+n=(\quad)$.
(A) 47
(B) 94
(C) 227
(D) 471
(E) 542
22. D.

From the problem, we know
$$
\begin{aligned}
P= & \frac{1}{(5!)^{3}}\left(\mathrm{C}_{5}^{1} \times 5(4!)^{3}-\mathrm{C}_{5}^{2} \times 5 \times 4(3!)^{3}+\right. \\
& \mathrm{C}_{5}^{3} \times 5 \times 4 \times 3(2!)^{3}-\mathrm{C}_{5}^{4} \times 5 \times 4 \times 3 \times \\
& \left.2(1!)^{3}+\mathrm{C}_{5}^{5} \times 5!\right) \\
& \frac{1}{(5!)^{3}}\left(5^{2}(4!)^{3}-10 \times 5 \times 4(3!)^{3}+\right. \\
& \left.10 \times 5 \times 4 \times 3(2!)^{3}-5 \times 5!+5!\right) \\
= & \frac{71}{400} .
\end{aligned}
$$

Therefore, $m+n=471$.