We wish to select 6 persons from 8, but if the person A is chosen, then B must be chosen. In how many ways can the selections be made?

Options

A. 20
B. 21
C. 22
D. 30

Explanation

To solve the problem of selecting 6 persons from 8 with the condition that if person A is chosen, then person B must also be chosen, we can break down the solution into clear steps.

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to select 6 persons from a total of 8. The key condition is that if person A is included in the selection, then person B must also be included.

2. **Case Analysis**: We will analyze two cases based on whether person A is selected or not.

- **Case 1**: Person A is selected.
- **Case 2**: Person A is not selected.

3. **Case 1: Person A is Selected**:
- If person A is selected, then person B must also be selected. This means we have already selected 2 persons (A and B).
- We now need to select 4 more persons from the remaining 6 persons (since A and B are already chosen).
- The number of ways to select 4 persons from 6 is given by the combination formula \( \binom{n}{r} \), which is \( \binom{6}{4} \).

\[ \text{Number of ways} = \binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6!}{4! \cdot 2!} = \frac{6 \times 5}{2 \times 1} = 15 \]

4. **Case 2: Person A is Not Selected**:
- If person A is not selected, then there are 7 remaining persons (B and 6 others).
- We need to select all 6 persons from these 7.
- The number of ways to select 6 persons from 7 is given by \( \binom{7}{6} \).

\[ \text{Number of ways} = \binom{7}{6} = \frac{7!}{6!(7-6)!} = \frac{7!}{6! \cdot 1!} = 7 \]

5. **Total Number of Ways**: Now, we add the number of ways from both cases to get the total number of selections.

\[ \text{Total ways} = \text{Ways from Case 1} + \text{Ways from Case 2} = 15 + 7 = 22 \]

### Final Answer:
The total number of ways to select 6 persons from 8, given the condition, is **22**.

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