imc

Let our denominator be $(5!{)}^3$, so we consider all possible distributions. We use PIE (Principle of Inclusion and Exclusion) to count the successful ones. When we have at $1$ box with all $3$ balls the same color in that box, there are ${}_5{C}_1{\cdot }_5{P}_1\cdot (4!{)}^3$ ways for the distributions to occur (${}_5{C}_1$ for selecting one of the five boxes for a uniform color, ${}_5{P}_1$ for choosing the color for that box, $4!$ for each of the three people to place their remaining items). However, we overcounted those distributions where two boxes had uniform color, and there are ${}_5{C}_2{\cdot }_5{P}_2\cdot (3!{)}^3$ ways for the distributions to occur (${}_5{C}_2$ for selecting two of the five boxes for a uniform color, ${}_5{P}_2$ for choosing the color for those boxes, $3!$ for each of the three people to place their remaining items). Again, we need to re-add back in the distributions with three boxes of uniform color... and so on so forth. Our success by PIE is $${}_5{C}_1{\cdot }_5{P}_1\cdot (4!{)}^3{-}_5{C}_2{\cdot }_5{P}_2\cdot (3!{)}^3{+}_5{C}_3{\cdot }_5{P}_3\cdot (2!{)}^3{-}_5{C}_4{\cdot }_5{P}_4\cdot (1!{)}^3{+}_5{C}_5{\cdot }_5{P}_5\cdot (0!{)}^3=120\cdot 2556.$$ $$\frac{120\cdot 2556}{120^3}=\frac{71}{400},$$ yielding an answer of $\boxed{471}$.