# Difficult Permutation-Combination Solved QuestionAptitude Discussion

 Q. For a scholarship, at most n candidates out of $2n + 1$ can be selected. If the number of different ways of selection at least one candidate is 63, the maximum number of candidates that can be selected for the scholarship is:
 ✖ A. 2 ✔ B. 3 ✖ C. 4 ✖ D. 5

Solution:
Option(B) is correct

The number of ways of selecting at least one candidate from $2n+1$ candidates (with a maximum number of candidates being selected is $n$) is,

$^{2n+1}C_1+^{2n+1}C_2+ ... +^{2n+1}C_n=63$

Now, take each answer choice for $n$ and substitute and check for which value of $n$, the above equation is satisï¬ed.

If we take choice (2), we have $n = 3$.

This means that $2n+1 = 7$.

If the value of $^7C_1+^7C_2+^7C_3$ is 63, it means that $n = 3$ will be the correct answer.

$=^7C_1+^7C_2+^7C_3\\ = 7 \left(7 \times \dfrac{6}{2}\right)+\left(7 \times \dfrac{6}{2}\times \dfrac{5}{3}\right)\\ =7+21+35\\ =\textbf{63}$

## (2) Comment(s)

Ajay
()

For example, if we consider $3n$ persons we get $7(2^n-1)$ no of ways with at least one person by the formula $2^n-1$.

If, I use this $2^n-1$ formula, I get $2.5$ as my answer why it differs?

ABHIJEET
()

thats because u are calculating for 2n+1 student.. in the question it is asked for just n students