10) show that in a group of 10 people (where any two people are either friends or enemies), there are either three mutual friends or four mutual enemies, and there are either three mutual enemies or four mutual friends.

Answer with explanation:

Number of People in the group =10 People

The combination between two people is that, they can be either friends or enemies.

Total number of Possible Relation

                            =18 +16+14+12+10+8+6+4+2

                            = 90 Relations in all.

Out of 90 relations , 45 will be friends and 45 will be enemies.

⇒Now, we have to prove that, between 10 people, there are either three mutual friends or four mutual enemies, and there are either three mutual enemies or four mutual friends.

→→If there are three mutual friends, total number of people in the group =3 ×2=6 people in the group

And, four mutual enemies in a group means there are 2 people in each group.

So, total number of people if we combine the two groups in which there are either three mutual friends or four mutual enemies

   =6 +4

   =10

Hence proved.

→→→→Second part is ,in this group there can be either three mutual enemies or four mutual friends.

⇒If there are three mutual enemies, total number of people in the group =3 ×2=6 people in the group

And, four mutual friends in a group means there are 2 people in each group.

So, total number of people if we combine the two groups in which there are either three mutual enemies or four mutual enemies

   =6 +4

   =10

Hence proved.

it could 14.28 but if its not that one then it has to be the 17 one. hope this ! : )

step-by-step explanation: