A person has 12 friends out of which 7 are relatives. In how many ways can he invite 6 friends such that at least 4 of them are relatives?

Options

A. 462
B. 562
C. 450
D. 400

Answer

To solve the problem of how many ways a person can invite 6 friends such that at least 4 of them are relatives, we can break it down into cases based on the number of relatives invited.

### Step-by-Step Solution:

1. **Identify the total number of friends and relatives**:
- Total friends = 12
- Relatives = 7
- Non-relatives = 12 - 7 = 5

2. **Define the cases based on the number of relatives invited**:
- We need to consider three cases for inviting at least 4 relatives:
- Case 1: 4 relatives and 2 non-relatives
- Case 2: 5 relatives and 1 non-relative
- Case 3: 6 relatives and 0 non-relatives

3. **Calculate the number of ways for each case**:
- **Case 1**: 4 relatives and 2 non-relatives
- Number of ways to choose 4 relatives from 7: \( \binom{7}{4} \)
- Number of ways to choose 2 non-relatives from 5: \( \binom{5}{2} \)
- Total for Case 1: \( \binom{7}{4} \times \binom{5}{2} \)

- **Case 2**: 5 relatives and 1 non-relative
- Number of ways to choose 5 relatives from 7: \( \binom{7}{5} \)
- Number of ways to choose 1 non-relative from 5: \( \binom{5}{1} \)
- Total for Case 2: \( \binom{7}{5} \times \binom{5}{1} \)

- **Case 3**: 6 relatives and 0 non-relatives
- Number of ways to choose 6 relatives from 7: \( \binom{7}{6} \)
- Number of ways to choose 0 non-relatives from 5: \( \binom{5}{0} \)
- Total for Case 3: \( \binom{7}{6} \times \binom{5}{0} \)

4. **Calculate the combinations**:
- For Case 1:
\[ \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]
Total for Case 1: \( 35 \times 10 = 350 \)

- For Case 2:
\[ \binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7 \times 6}{2 \times 1} = 21 \]
\[ \binom{5}{1} = 5 \]
Total for Case 2: \( 21 \times 5 = 105 \)

- For Case 3:
\[ \binom{7}{6} = 7 \]
\[ \binom{5}{0} = 1 \]
Total for Case 3: \( 7 \times 1 = 7 \)

5. **Add the totals from all cases**:
\[ \text{Total ways} = 350 + 105 + 7 = 462 \]

### Final Answer:
The total number of ways the person can invite 6 friends such that at least 4 of them are relatives is **462**.