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Chapter 1: Problem 20

A person has 8 friends, of whom 5 will be invited to a party. (a) How many choices are there if 2 of the friends are feuding and will not attend together? (b) How many choices if 2 of the friends will only attend together?

Short Answer

Expert verified
(a) 16 choices if the feuding friends cannot both be invited. (b) 35 choices if the two friends will only attend together.

Step by step solution

01

(a) Feuding friends case:

First, let's tackle the case of the feuding friends. In this case, we have 3 possible scenarios: 1. We can choose to invite neither of the feuding friends. 2. We invite one of the feuding friends but not the other. 3. We invite the other feuding friend but not the first. We will find the number of ways of inviting friends in each of these scenarios and then combine the results.
02

Scenario 1: Neither feuding friend is invited:

In this case, we have to select 5 friends from the remaining 6 (8 total friends - 2 feuding friends). This is a simple combination problem, and the number of ways to invite 5 friends from 6 can be calculated as: \(\binom{6}{5} = \frac{6!}{5!(6-5)!} = 6\)
03

Scenario 2: One feuding friend is invited:

Without loss of generality, let's assume we invite the first feuding friend but not the second. In this case, we have to select the remaining 4 friends out of the 5 non-feuding friends (7 total friends - the uninvited feuding friend). \(\binom{5}{4} = \frac{5!}{4!(5-4)!} = 5\)
04

Scenario 3: The other feuding friend is invited:

This scenario is symmetrical to the second scenario. We invite the second feuding friend but not the first, and we still have to choose the remaining 4 friends out of the 5 non-feuding friends. \(\binom{5}{4} = \frac{5!}{4!(5-4)!} = 5\)
05

Total count of (a):

To find the total number of ways for part (a), we take the sum of the number of ways from all the scenarios: \(6+5+5 = 16\) Thus, there are 16 different ways to invite friends without inviting both feuding friends.
06

(b) Friends attending together only:

Now, let's consider the given condition that there are 2 friends who will attend the party only if they are invited together. Let's treat this pair as a single entity. In this case, we are selecting 4 guests out of 7 entities (6 individual friends and 1 "pair" of friends). The number of ways to choose 4 from the 7 can be calculated as: \(\binom{7}{4} = \frac{7!}{4!(7-4)!} = 35\) Thus, there are 35 different ways to invite friends with the condition that the two friends will attend the party only if they are invited together. The solutions to the problem are: (a) 16 choices if the feuding friends cannot both be invited. (b) 35 choices if the two friends will only attend together.

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