Three officers – a president, a treasurer, and a secretary – are to be chosen from among four people: Alice, Bob, Cyd, and Dan Three officers – a president, a treasurer, and a secretary – are to be chosen from among four people: Alice, Bob, Cyd, and Dan How many ways can the officers be chosen if Bob is not qualified to be treasurer and Cyd is not qualified to be secretary? Answer given: By inclusion-exclusion principle, $24−6−6 + 2 = 14$ But I am not sure how to solve it.

Let $A$ be the set of all officer assignments with no restrictions. Note that $| A | = 4 \cdot 3 \cdot 2 = 24$.

Let $B$ be the set of all officer assignments such that Bob is treasurer, and let $C$ be the set of all officer assignments such that Cyd is secretary. Note that $|B| = |C| = 3 \cdot 2 = 6$.

The number of officer assignments such that Bob is not treasurer and Cyd is not secretary is $$ |A| - | B \cup C| = 24 - (\underbrace{|B| + |C| - |B \cap C|}_{\text{inclusion-exclusion formula}}) = 24 - (6 + 6 - 2) = 14. $$