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.
Best Answer
First you find the total permutations of $4$ candidates chosen $3$ at a time: $$P(4,3)=\frac{4!}{(4-3)!}=24.$$ Then you subtract the choices when Bob is a treasurer: $$P(3,2)=\frac{3!}{(3-2)!}=6.$$ Then you subtract the choices when Cyd is a secretary: $$P(3,2)=\frac{3!}{(3-2)!}=6.$$ Then you add the choices when Bob is a treasurer as well as Cyd is a secretary: $$P(2,1)=\frac{2!}{(2-1)!}=2.$$