A) \[{{n}^{p}}\]
B) \[{{p}^{n}}\]
C) \[{{(p+1)}^{n}}-1\]
D) \[{{(n+1)}^{p}}-1\]
Correct Answer: C
Solution :
In case of each book we may take 0, 1, 2, 3,...,p copies, i.e., we may deal with each book in \[(p+1)\]ways and therefore with all the books in\[{{(p+1)}^{n}}\] ways. But this includes the case where all the books rejected and no selection is made. So, the number of ways in which selection can be made \[={{(p+1)}^{n}}-1\]
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