In an election, a voter may vote for any number of candidates not greater than the number to be chosen. There are 7 candidates and 4 members are to be chosen. In how many ways can a person vote?

Options

A. 89
B. 98
C. 79
D. 101

Answer

To solve the problem, we need to determine the number of ways a voter can vote for candidates in an election where there are 7 candidates and a voter can choose up to 4 candidates. The voter can choose to vote for 1, 2, 3, or 4 candidates.

### Step-by-Step Solution:

1. **Identify the Candidates and Choices**:
- There are 7 candidates.
- A voter can choose to vote for 1, 2, 3, or 4 candidates.

2. **Calculate the Combinations for Each Case**:
- The number of ways to choose \( r \) candidates from \( n \) candidates is given by the combination formula \( nCr = \frac{n!}{r!(n-r)!} \).

3. **Calculate for 1 Candidate**:
- For choosing 1 candidate from 7:
\[ 7C1 = \frac{7!}{1!(7-1)!} = \frac{7!}{1! \cdot 6!} = 7 \]

4. **Calculate for 2 Candidates**:
- For choosing 2 candidates from 7:
\[ 7C2 = \frac{7!}{2!(7-2)!} = \frac{7!}{2! \cdot 5!} = \frac{7 \times 6}{2 \times 1} = 21 \]

5. **Calculate for 3 Candidates**:
- For choosing 3 candidates from 7:
\[ 7C3 = \frac{7!}{3!(7-3)!} = \frac{7!}{3! \cdot 4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

6. **Calculate for 4 Candidates**:
- For choosing 4 candidates from 7:
\[ 7C4 = \frac{7!}{4!(7-4)!} = \frac{7!}{4! \cdot 3!} = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} = 35 \]

7. **Sum the Combinations**:
- Now, we add the number of ways to vote for 1, 2, 3, and 4 candidates:
\[ \text{Total Ways} = 7C1 + 7C2 + 7C3 + 7C4 = 7 + 21 + 35 + 35 = 98 \]

### Final Answer:
The total number of ways a person can vote is **98**.

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