In organizing groups to work on homework together every student is asked to fill out a form listing all other students who they would be willing to work with. If there are 251 students in the class and each student lists exactly 168 other students who they might be willing to work with. For any two students in the class, if student A puts student B on their list, then students B will also have student A on their list.

Required:
Using the Pidgeon Hole show that there must be some group of four students who are work with one another.

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ogorwyne ogorwyne · Answer 1 · 2022-04-05T00:50:53+02:00

The principle shows that there would be some group of 4 students that are willing to work with one another.

The goal here is to help the students to form homework groups using the Pigeon hole system.

The number of students in A list is 168, this is 167 when we subtract B.

168 students have A in their list. This is 167 when we subtract A.

Total students = 250-1 ( We have to subtract A and B) = 249

249 - 167 = 82 students have B in their list and not A.

167 - 82 = 85, these are the students that have both A and B in their list.

We have a student C who have both A and B in the lis.

Thre are 84 students other than c who have A and B in the list. This is a set called s.

168 students have c in their list. There have to be at least 2 students from s in this number.

This would leave 166 students other than s.

At least 1 student wants c, a and b in the list.

We conclude that there is one group of 4 who are to work with one another.

In mathematics, this principle states that if objects are in a container, then at least one container is going to have more than one item.

Read more on pigeon hole principle here: https://brainly.com/question/7455939