Question

In an election, number of candidates excess the number to be elected by $2$. A man can vote $56$ ways. Then, the number of candidates is

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

Answer: (B)

$6$

Let number of candidates be $n$

Therefore, $n$ are to be elected and so one can vote up $(n-2)$.

Hence, number of ways in which one can vote
${=}^n{C}_1{+}^n{C}_2+....{+}^n{C}_{n-2}=56$ (given)
$\Rightarrow 2^n-{(}^n{C}_0{+}^n{C}_{n-1}{+}^n{C}_n)=56$
$\Rightarrow 2^n-n=58\Rightarrow 2^n=58+n$
Which is satisfied by $n=6$ only.