There are $P$ copies of $n$-different books. The number of different ways in which a non-empty selection can be made from them, is

Number of selections of any number of copies of a book $=P+1$, (because copies of the same book are identical things). Similarly, for each book. $\therefore$ Total number of selections $\begin{aligned} & =(P+1)(P+1) \ldots \text { to } n \text { factors } \\ & =(P+1)^n \end{aligned}$ But this includes a selections which is empty $i e$, zero copy of each is selected. Excluding this, the required number of non-empty selections $=(P+1)^n-1$