A student is allowed to select at least (n+1) | vedclass

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(B) The total number of ways to select at least $(n+1)$ books from $(2n+1)$ books is given by the sum of combinations:
$^{2n+1}C_{n+1} + ^{2n+1}C_{n+

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) The total number of ways to select at least $(n+1)$ books from $(2n+1)$ books is given by the sum of combinations:
$^{2n+1}C_{n+1} + ^{2n+1}C_{n+2} + \dots + ^{2n+1}C_{2n} = 255$.
Note that the student cannot select all books,so the term $^{2n+1}C_{2n+1}$ is excluded.
We know that the sum of all combinations from $^{2n+1}C_0$ to $^{2n+1}C_{2n+1}$ is $2^{2n+1}$.
Since $^{2n+1}C_k = ^{2n+1}C_{2n+1-k}$,the sum of the first half of the terms is equal to the sum of the second half.
Specifically,$\sum_{k=n+1}^{2n} {^{2n+1}C_k} = \frac{2^{2n+1}}{2} - 1 = 2^{2n} - 1$.
Given $2^{2n} - 1 = 255$,we have $2^{2n} = 256$.
Since $256 = 2^8$,we get $2n = 8$,which implies $n = 4$.
The total number of books is $2n + 1 = 2(4) + 1 = 9$.

$A$ student is allowed to select at least $(n+1)$ books but not all books from a collection of $(2n+1)$ books. If the total number of ways in which he can select these books is $255$,then the number of books in that collection is

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