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?

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

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Asked at: Optiver, SIG

A guest list with a feud

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