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adell [148]
3 years ago
15

10) Show that in a group of 10 people (where any two people are either friends or enemies), there are either three mutual friend

s or four mutual enemies, and there are either three mutual enemies or four mutual friends.
Mathematics
1 answer:
ddd [48]3 years ago
3 0

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.

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