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.
On
As stated in the answer, you need to compute the total case without exclusion (assume anyone can take any seat), then subtract number of cases for each exclusion (Bob can't be treasurer, Cyd can't be secretary), then add back the number of cases for both Bob and Cyd exclusions (Bob can't be treasurer AND Cyd can't be secretary) where the intersection was doubly-subtracted previously.
On
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: $$P(2,1)=\frac{2!}{(2-1)!}=2.$$
On
Suppose $P,T,S$ are posts and $A,B,C,D$ are applicants. When there is no restriction, total possibilities are $4\times 3\times 2\times 1=24.$ Remove when $B$ is at post $T$.
Total such cases are: $3\times 2 \times 1$
when $C$ is at post $S$ total such cases are $3\times 2 \times 1$.
Following two cases are common in removal $PTS=ABC, DBC$ and has been removed twice so add $2.$
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 $$ |A| - | B \cup C| = 24 - (\underbrace{|B| + |C| - |B \cap C|}_{\text{inclusion-exclusion formula}}) = 24 - (6 + 6 - 2) = 14. $$