A guest list with a feud

EasyCombinatorics~3m

Jimmy will invite exactly $5$ of his $8$ friends to a party. But two particular friends are feuding and refuse to attend together. How many valid guest lists are possible?

Show hints (2)+

All lists: $\binom{8}{5}=56$ . Now remove those containing both feuding friends.
Both-in lists need $3$ more from the other $6$ : $\binom{6}{3}=20$ .

Answer

Reveal answer →

Create a free account to see the full worked solution and track your progress.

Create a free account

Interview format: Numerical Reasoning Tests, Betting Game Rounds

A guest list with a feud

Answer

Related questions