A lady gives a dinner party to 5 guests to be selected from nine friends. The number of ways of forming the party of 5, given that two of the friends will not attend the party together, is

Options

A. 56
B. 126
C. 91
D. None of these

Explanation

To solve the problem of selecting 5 guests from 9 friends, given that 2 specific friends will not attend the party together, we can break the solution down into several steps.

### Step-by-Step Solution:

1. **Identify the Total Friends and Constraints**:
- We have a total of 9 friends.
- Let’s denote the two friends who cannot attend together as A and B.

2. **Calculate the Total Combinations Without Constraints**:
- If there were no restrictions, the number of ways to choose 5 guests from 9 friends would be calculated using the combination formula:
\[ \text{Total ways} = \binom{9}{5} = \frac{9!}{5!(9-5)!} = \frac{9!}{5!4!} \]

3. **Calculate the Combinations When Both A and B Are Excluded**:
- If both A and B are excluded, we can choose all 5 guests from the remaining 7 friends:
\[ \text{Ways excluding A and B} = \binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!} = 21 \]

4. **Calculate the Combinations When One of A or B Is Included**:
- If one of A or B is included, we can choose 4 more guests from the remaining 7 friends. We can do this in two scenarios:
- Case 1: A is included, B is not.
- Case 2: B is included, A is not.
- For each case, we have:
\[ \text{Ways to choose 4 from 7} = \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = 35 \]
- Since there are 2 cases (A included or B included), we multiply by 2:
\[ \text{Total ways with one of A or B} = 2 \times 35 = 70 \]

5. **Combine Both Scenarios**:
- Finally, we add the two scenarios together to get the total number of ways to select the guests:
\[ \text{Total ways} = \text{Ways excluding A and B} + \text{Ways with one of A or B} = 21 + 70 = 91 \]

### Final Answer:
The total number of ways to form the party of 5 guests, given that A and B will not attend together, is **91 ways**.

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