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

Step-by-step solution

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. Total number of selections aligned & =(P+1)(P+1) to n factors & =(P+1)^n 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