A person has 8 friends, of whom 5 will be invited to a party. (a) How many different ways can the 5 invitations be sent if 2 of the friends are feuding and must not be invited at the same time? (b) How many different ways can the 5 invitations be sent if 2 of the friends must either both be invited or both not invited?

Make sure to recheck your answer such that you are 100% certain that it is correct

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:

\[
\boxed{30}
\]

---

### 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:

\[
\boxed{26}
\]

---

### 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.

A person has 8 friends, of whom 5 will be invited to a party. (a) How many different ways can the 5 invitations be sent if 2 of the friends are feuding and must not be invited at the same time? (b) How many different ways can the 5 invitations be sent if 2 of the friends must either both be invited or both not invited?

Explore how to solve a combinatorial problem involving invitations with constraints. Learn how to calculate the number of ways to invite friends to a party while accounting for specific conditions, using the combination formula. This problem is ideal for Grades 10-12 and covers key concepts in probability and combinatorics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation