In a n election, the number of candidates exceeds the number to be elected y 2. A man can vote in 56 ways. Find the number of candidates.

Explanation

To solve the problem step by step, we can follow this approach:

### Step 1: Define Variables
Let the number of candidates be \( n \). According to the problem, the number of candidates exceeds the number to be elected by 2. Therefore, if \( y \) is the number of candidates to be elected, we can express this as:
\[ n = y + 2 \]

### Step 2: Express Voting Ways
A man can vote in \( 56 \) ways. The ways a man can vote can be represented as the sum of combinations of selecting candidates from \( n \):
\[ \text{Ways to vote} = \binom{n}{1} + \binom{n}{2} + \binom{n}{3} + \ldots + \binom{n}{y} \]
Since \( y = n - 2 \), we can rewrite this as:
\[ \text{Ways to vote} = \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n-2} \]

### Step 3: Use the Binomial Theorem
We know from the Binomial Theorem that:
\[ \sum_{k=0}^{n} \binom{n}{k} = 2^n \]
This means:
\[ \sum_{k=0}^{n} \binom{n}{k} = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n} = 2^n \]
To find the sum from \( \binom{n}{1} \) to \( \binom{n}{n-2} \), we can subtract the terms \( \binom{n}{0} \), \( \binom{n}{n-1} \), and \( \binom{n}{n} \):
\[ \sum_{k=1}^{n-2} \binom{n}{k} = 2^n - \left( \binom{n}{0} + \binom{n}{n-1} + \binom{n}{n} \right) \]
This simplifies to:
\[ \sum_{k=1}^{n-2} \binom{n}{k} = 2^n - 1 - n - 1 = 2^n - n - 2 \]

### Step 4: Set Up the Equation
According to the problem, this sum equals \( 56 \):
\[ 2^n - n - 2 = 56 \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ 2^n - n = 58 \]

### Step 6: Trial and Error to Find \( n \)
We can now use trial and error to find \( n \):
- For \( n = 5 \):
\[ 2^5 - 5 = 32 - 5 = 27 \quad (\text{not a solution}) \]
- For \( n = 6 \):
\[ 2^6 - 6 = 64 - 6 = 58 \quad (\text{this works!}) \]
- For \( n = 7 \):
\[ 2^7 - 7 = 128 - 7 = 121 \quad (\text{too high}) \]

### Conclusion
Thus, the number of candidates \( n \) is \( 6 \).