Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to

$3 \mathrm{Shelf} \mathrm{empty}: (8,0,0,0) \to 1 \mathrm{way}$

$\left.2\text{ shelf empty: }\begin{array}{c}(7,1,0,0)\\ (6,2,0,0)\\ (5,3,0,0)\\ (4,4,0,0)\end{array}\right]\to \text{ 4ways }$

$\left.1\text{ shelf empty: }\begin{array}{cc}(6,1,1,0)& (3,2,1,2,0)\\ (4,3,1,0)& (4,2,2,0)\end{array}\right]\to 5\text{ ways }$

$\left.0\text{ Shelf empty : }\begin{array}{ll}(1,2,3,2)& (5,1,1,1)\\ (2,2,2,2)& \\ (3,3,1,1)& \\ (4,2,1,1)& \end{array}\right]\to \text{ 5ways }$

Total = 15 ways