Every ball has two options

$\Rightarrow 4$ balls can be put in $2^4$ ways.

Arrangement among themselves is $2!$

So, total ways$=\frac{1}{2!}\times 2^4= 2^3$

But, above count also includes the one case in which all the balls are put in one box,

$\Rightarrow$ required number of ways  $k= 2^3-1=7$