Question:medium

In an election, the number of candidates is one more than the number of seats. If a voter can cast his vote in 30 ways, find the number of candidates (when a voter can cast his vote for one or more seats).

Show Hint

A frequently used identity: \[ \sum_{r=1}^{n-1} {n \choose r} = 2^n-2 \] obtained by removing the empty set and the whole set from all subsets.
Updated On: Jun 16, 2026
  • \(31\)
  • \(29\)
  • \(5\)
  • \(6\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Name the unknown.
Let the number of candidates be $n$. Then the number of seats is $n-1$, one fewer than the candidates.

Step 2: Understand a single vote.
A voter may vote for one or more candidates, but cannot vote for more than the number of seats, which is $n-1$. So a voter picks any group of size $1$ up to $n-1$.

Step 3: Count the voting groups.
The number of ways is $\binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n-1}$.

Step 4: Use the full-subset total.
We know $\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = 2^n$. Removing the two end terms $\binom{n}{0} = 1$ (vote for nobody) and $\binom{n}{n} = 1$ (vote for everyone, too many) leaves $2^n - 2$.

Step 5: Set it equal to 30.
So $2^n - 2 = 30$, giving $2^n = 32$, hence $n = 5$.

Step 6: State the answer.
The number of candidates is $5$. (As a check, with $5$ candidates the voter chooses groups of size $1$ to $4$, totalling $5 + 10 + 10 + 5 = 30$.) \[ \boxed{5} \]
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