Question

If the number of ways in which four distinct balls can be put into two identical boxes so that no box remains empty is equal to $k$, then $k$ is

Answer 1

Every ball has two options
$\Rightarrow 4$ balls can be put in $2^4$ ways.
As given that boxes are identical so,
arrangement is $=\frac{1}{2!}\times 2^4=2^3$
But, above count also includes the one case in which all the balls are put in one box.
$\Rightarrow$ Number of ways $=2^3-1=7$

Alternate Solution :
We can divide the balls into two groups : $(1,3)$ and $(2,2)$
In $(2,2),$ both groups have equal number of balls.
Now, number of ways
$=\frac{4!}{1!3!}+\frac{4!}{2!2!}\times \frac{1}{2!}$
$=4+3=7$