Counting

Because both the balls and the boxes are indistingusable, we only have to account for each case once.

Here are the only possibilites: 

$4$ in one basket and $0$ in the rest.

$3$ in one basket and $1$ in $1$ of the remaining $4$ baskets, with none in the rest

$2$ in one basket and $2$ in another basket, with none in the rest

$2$ in one basket and $1$ in another and $1$ more in another basket, with none in the rest

$1$ in one basket and $3$ in another, with none in the rest.

This accounts for a total of $\color{brown}\boxed5$ ways.