Number of ways of arranging 8 identical books into 4 identical...

We have $4$ identical shelves in which any number of shelves can be empty.

So, we will put $3, 2, 1, 0$ shelves empty one by one.

Also, the non-empty shelves won't require internal arrangement as all the shelves are identical.

$3$ Shelf empty: \left(8, 0, 0, 0\right) \to 1$way</p><p>$2shelf empty: \left$\begin{array}{ll}( 7, 1, 0, 0 ) \\ ( 6, 2, 0, 0 ) \\ ( 5, 3, 0, 0 ) \\ ( 4, 4, 0, 0 )\end{array}$\right] \to 4$ ways</p><p>$1$ shelf empty: \left\begin{cases}( 6, 1, 1, 0 ) & ( 3, 3, 2, 0 ) \\ ( 5, 2, 1, 0 ) & ( 4, 2, 2, 0 ) \\ ( 4, 3, 1, 0 ) & \end{cases}\right] \to 5$ways</p><p>$0shelf empty: \left\begin{cases}( 1, 2, 3, 2 ) & ( 5, 1, 1, 1 ) \\ ( 2, 2, 2, 2 ) & \\ ( 3, 3, 1, 1 ) & \\ ( 4, 2, 1, 1 ) & \end{cases}\right] \to 5 ways

Total $= 15$ ways