Question

In how many ways can n different objects put in 3 different boxes so that no one remains empty?

Answer 1

We know that

The number of ways in which $(m+n+p)$ differnt things can be divided into three different groiups containing m,n, and p is $\frac{(m+n+p)!}{m!n!p!}$

But here $m+n+p=n$

Hence no of ways

$\frac{n!}{{\left(\frac{n!}{3}\right)}^3}$