Distribute the balls

[tanmoy]

Find the number of distributions of five red balls and five blue balls into three distinct boxes,with no empty boxes.

[Fm Jakaria]

We denote by $S(n)$ to be the number of ways $5$ balls of same colour can be kept in $n$ boxes, with possibly some boxes empty. Now note that $S(n)^2$ is the number of ways five red and five blue balls can be kept in $n$ boxes in this way; considering disjoint events. So our desired result is $S(3)^2-(^3C_1)S(2)^2+(^3C_2)S(1)^2$, considering cases where at least $0,1$ boxes are empty, and considering overcouting arising from exacly two empty boxes; while deciding which boxes are empty. And computing S(n) is the same as counting ordered n-tuple of nonngegative integers $a_i$ that sum to $5$; and its exacly same as number of ways putting $n-1$ equivalent plus signs between and edges of n dots, where multiple pluses are allowed in a place. Considering cases of all multiplicity of pluses:
$S(3) = ^6C_2+^6C_1$, $S(2) = ^6C_1$, $S(1) = 1$.

[tanmoy]

We denote by $S(n)$ to be the number of ways $5$ balls of same colour can be kept in $n$ boxes, with possibly some boxes empty. Now note that $S(n)^2$ is the number of ways five red and five blue balls can be kept in $n$ boxes in this way; considering disjoint events. So our desired result is $S(3)^2-(^3C_1)S(2)^2+(^3C_2)S(1)^2$, considering cases where at least $0,1$ boxes are empty, and considering overcouting arising from exacly two empty boxes; while deciding which boxes are empty. And computing S(n) is the same as counting ordered n-tuple of nonngegative integers $a_i$ that sum to $5$; and its exacly same as number of ways putting $n-1$ equivalent plus signs between and edges of n dots, where multiple pluses are allowed in a place. Considering cases of all multiplicity of pluses:
$S(3) = ^6C_2+^6C_1$, $S(2) = ^6C_1$, $S(1) = 1$.

Nice solution.Thank you