At 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 ?

Answer

To solve the problem of how many ways a person can vote for candidates in an election where there are 7 candidates and 4 members to be chosen, we can break it down step by step.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have 7 candidates.
- We need to choose up to 4 members.
- A voter can choose any number of candidates from 0 to 4.

2. **Choosing Candidates**:
- The possible choices for the number of candidates a voter can select are:
- Choosing 0 candidates (not voting at all)
- Choosing 1 candidate
- Choosing 2 candidates
- Choosing 3 candidates
- Choosing 4 candidates

3. **Using Combinations**:
- The number of ways to choose \( r \) candidates from \( n \) candidates is given by the combination formula \( \binom{n}{r} \).
- Therefore, we need to calculate:
- \( \binom{7}{0} \) for choosing 0 candidates
- \( \binom{7}{1} \) for choosing 1 candidate
- \( \binom{7}{2} \) for choosing 2 candidates
- \( \binom{7}{3} \) for choosing 3 candidates
- \( \binom{7}{4} \) for choosing 4 candidates

4. **Calculating Each Combination**:
- \( \binom{7}{0} = 1 \) (There is 1 way to choose no candidates)
- \( \binom{7}{1} = 7 \) (There are 7 ways to choose 1 candidate)
- \( \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 \) (There are 21 ways to choose 2 candidates)
- \( \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \) (There are 35 ways to choose 3 candidates)
- \( \binom{7}{4} = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} = 35 \) (There are 35 ways to choose 4 candidates)

5. **Summing All the Combinations**:
- Now, we add all the combinations together:
\[ \text{Total ways} = \binom{7}{0} + \binom{7}{1} + \binom{7}{2} + \binom{7}{3} + \binom{7}{4} \]
\[ = 1 + 7 + 21 + 35 + 35 = 99 \]

### Final Answer:
Thus, the total number of ways a person can vote is **99**.