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Answer

n = 3

Explanation

Since, student is allowed to select at most n books out of (2n + 1) books.
`therefore" " ""^(2n + 1)C_(1) + ""^(2n + 1)C_(2)+.....+ ""^(2n + 1)C_(n) = 63 " "...(i)`
We know `""^(2n + 1)C_(0) + ""^(2n + 1)C_(1)+.....+ ""^(2n + 1)C_(2n + 1) = 2^(2n + 1)`
`rArr 2(""^(2n + 1)C_(0) + ""^(2n + 1)C_(1)+ ""^(2n + 1)C_(2) + ... + ""^(2n + 1)C_(n)) = 2^(2n + 1)`
`rArr " "^(2n + 1)C_(1)+ ""^(2n + 1)C_(2) + ... + ""^(2n + 1)C_(n) = (2^(2n - 1))" ....(ii)"`
From Eqs. (i) and (ii), we get
`2^(2n) - 1 = 63`
`rArr" " 2^(2n) = 64`
`rArr " " 2n = 6`
`rArr" " n = 3`