Q88 / 200 · Manipal (MET) 2012 · Mathematics

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

  • A
    5
  • Correct answer
    B
    6
  • C
    7
  • D
    8

Solution

Let number of candidates be $n$. Therefore, $n-2$ are to be elected and so one can vote up to $(n-2)$. Hence, number of ways in which one can vote $\begin{aligned} & ={ }^n C_1+{ }^n C_2+\ldots+{ }^n C_{n-2}=56 \qquad \mathrm{(given)} \\ & \Rightarrow 2^n-\left({ }^n C_0+{ }^n C_{n-1}+{ }^n C_n\right)=56 \\ & \Rightarrow 2^n-n=58 \Rightarrow 2^n=58+n \end{aligned}$
Which is satisfied by $n=6$ only.

Hence, the correct answer is option: 'B'.

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