Question

There are three boxes $A,B$ and $C$ having capacities of $5,2$ and $2$ balls respectively then the number of ways of putting $7$ different balls into the boxes such that no box remains empty is

Answer 1

There is no way that box $A$ remains empty so there is chance that box $B$ or box $C$ remains empty

Total ways to distribute balls in the boxes $A,B,C$ could be $(3,2,2)$, $(4,2,1)$, $(4,1,2)$ and $(5,1,1)$
So total ways will be
$={}^7{C}_3\times {}^4{C}_2\times {}^2{C}_2+{}^7{C}_4\times {}^3{C}_2\times {}^1{C}_1+{}^7{C}_4\times {}^3{C}_1\times {}^2{C}_2+{}^7{C}_5\times {}^2{C}_1\times {}^1{C}_1=462$