A persons has got 12 friends of whom 8 are relatives. In how many ways can he invite 7 guests such that 5 of them may be relatives ?

Answer

To solve the problem of inviting 7 guests such that 5 of them are relatives, we can break it down into steps:

### Step-by-Step Solution:

1. **Identify the Total Friends and Their Categories**:
- Total friends = 12
- Relatives = 8
- Non-relatives = 12 - 8 = 4

2. **Determine the Composition of the Guests**:
- We need to invite a total of 7 guests.
- Out of these, 5 must be relatives and the remaining 2 must be non-relatives.

3. **Calculate the Ways to Choose Relatives**:
- The number of ways to choose 5 relatives from 8 is given by the combination formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
- Here, \( n = 8 \) and \( r = 5 \):
\[ \binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!} \]

4. **Calculate the Ways to Choose Non-Relatives**:
- The number of ways to choose 2 non-relatives from 4 is:
\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} \]

5. **Calculate Each Combination**:
- For relatives:
\[ \binom{8}{5} = \frac{8 \times 7}{2 \times 1} = 28 \]
- For non-relatives:
\[ \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 \]

6. **Calculate the Total Number of Ways**:
- The total number of ways to invite the guests is the product of the two combinations:
\[ \text{Total Ways} = \binom{8}{5} \times \binom{4}{2} = 28 \times 6 = 168 \]

### Final Answer:
The total number of ways the person can invite 7 guests such that 5 of them are relatives is **168**.