Let's break this problem into two parts and ensure every step is clear and accurate:

Given:

• A person has 8 friends.
• 5 friends will be invited to a party.
• Special constraints are placed on two of these friends.

Part (a): Two of the friends are feuding and must not be invited at the same time.

Step-by-Step Approach:

1. Without Constraints: First, let's calculate the total number of ways to invite 5 friends from 8 without any restrictions. This is a combination problem, and the formula for combinations is:

$\text{Total ways} = \binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!}$

2. Feuding Friends Condition: Let’s assume the two feuding friends are $F_1$ and $F_2$. Since $F_1$ and $F_2$ must not both be invited, there are two cases to consider:

   • Case 1: Invite $F_1$ but not $F_2$.
   • Case 2: Invite $F_2$ but not $F_1$.

   In both cases, we still need to select 4 more friends from the remaining 6 friends (since one of $F_1$ or $F_2$ is already selected, leaving 6 other friends).

3. Ways for Each Case: The number of ways to invite 4 friends from 6 is:

$\binom{6}{4} = \frac{6!}{4!(6-4)!} = 15$

4. Total for Feuding Condition: Since there are 2 cases (either invite $F_1$ or invite $F_2$), the total number of ways to invite 5 friends such that $F_1$ and $F_2$ are not invited together is:

$2 \times \binom{6}{4} = 2 \times 15 = 30$

Thus, the answer to part (a) is:

$\boxed{30}$

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Part (b): Two of the friends must either both be invited or both not invited.

Step-by-Step Approach:

1. Without Constraints: Again, the total number of ways to invite 5 friends from 8 without any restrictions is:

$\binom{8}{5} = 56$

2. Friends Must Come Together: Let’s assume these two friends are $F_1$ and $F_2$, and they must either both be invited or both not be invited. There are two main cases:

   • Case 1: Both $F_1$ and $F_2$ are invited. In this case, we need to choose 3 more friends from the remaining 6 friends. The number of ways to do this is:

   $\binom{6}{3} = \frac{6!}{3!(6-3)!} = 20$

   • Case 2: Neither $F_1$ nor $F_2$ is invited. In this case, we need to choose all 5 friends from the remaining 6 friends. The number of ways to do this is:

   $\binom{6}{5} = \frac{6!}{5!(6-5)!} = 6$

3. Total for "Together" Condition: Adding the two cases together, we get the total number of ways to invite 5 friends such that $F_1$ and $F_2$ are either both invited or both not invited:

$\binom{6}{3} + \binom{6}{5} = 20 + 6 = 26$

Thus, the answer to part (b) is:

$\boxed{26}$

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Summary:

• (a) The number of ways to invite 5 friends such that two feuding friends are not invited together is 30.
• (b) The number of ways to invite 5 friends such that two friends are either both invited or both not invited is 26.

Would you like further details or explanations?

Here are some related questions:

1. How do you calculate combinations for larger sets?
2. What if there were 3 friends feuding instead of 2?
3. How would the answer change if 6 friends were to be invited instead of 5?
4. Can the combination formula be generalized for more than two specific friends?
5. How do constraints affect the complexity of counting problems?

Tip: When working with combinatorial problems involving constraints, always try breaking the problem into smaller cases or subproblems to avoid errors.