Question

A student is allowed to select at most n books from a lot of 2n+1 books. The number of ways in which he can select a book is 63, then the value of n equals

• A. 64
• B. 32
• C. 16
• D. None of these

Answer 1

The correct option is D None of these

The number of ways of selecting n books from 2n+1 will be
${}^{2n+1}{C}_1+{}^{2n+1}{C}_2+{}^{2n+1}{C}_3+{}^{2n+1}{C}_4+...{}^{2n+1}{C}_n=63$
${}^{2n+1}{C}_0+{}^{2n+1}{C}_1+{}^{2n+1}{C}_2+{}^{2n+1}{C}_3+{}^{2n+1}{C}_4+...{}^{2n+1}{C}_n=64$
Now
${}^{2n+1}{C}_0+{}^{2n+1}{C}_1+{}^{2n+1}{C}_2+{}^{2n+1}{C}_3+{}^{2n+1}{C}_4+...{}^{2n+1}{C}_{2n}=2^{2n+1}$
${}^{2n+1}{C}_r={}^{2n+1}{C}_{2n+1-r}$
Hence
$2({}^{2n+1}{C}_0+{}^{2n+1}{C}_1+{}^{2n+1}{C}_2+{}^{2n+1}{C}_3+{}^{2n+1}{C}_4+...{}^{2n+1}{C}_n)=2^{n+1}$
$({}^{2n+1}{C}_0+{}^{2n+1}{C}_1+{}^{2n+1}{C}_2+{}^{2n+1}{C}_3+{}^{2n+1}{C}_4+...{}^{2n+1}{C}_n)=2^{2n}$
Hence $2^{2n}=64$
$2n=6$
$n=3$