Choosing five guests when two cannot both come

Question

From a group of $8$ friends, exactly $5$ are to be invited to a party. Two particular friends will not come if the other is there, so they cannot both be invited. How many valid guest lists are there?

Short answer

$36$ . All $\binom85=56$ lists minus those containing both feuding friends, which are fixed by choosing $3$ of the other $6$ : $\binom63=20$ . So $56-20=36$ .

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Lessons behind this question

Counting and Combinatorics
Sample Spaces, Events, and the Probability Axioms
Conditional Probability, Total Probability, and Bayes' Rule

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