Combinatorics Through Guided Discovery, 2004a

96 5. The Principle of Inclusion and Exclusion
Problem 232. Give a proof of your formula for the principle of inclusion
and exclusion. (h)
Problem 233. We get a more elegant proof if we ask for a picture enumerator
for A 1 ∪ A 2 ∪···∪A n . so let us assume A is a set with a picture function P
defined on it and that each set A i is a subset of A.
(a) By thinking about how we got the formula for the size of a union,
write down instead a conjecture for the picture enumerator of a union.
You could use notation like E P ( ⋂ i:i∈S A i ) for the picture enumerator
of the intersection of the sets A i for i in a subset of S of [n].
(b) If x ∈ ⋃ n
i=1 A i, what is the coefficient for P(x) in (the inclusionexclusion
side of) your formula for E P ( ⋃ n
i=1 A i)? (h)
(c) If x ⋃ n
i=1 A i, what is the coefficient of P(x) in (the inclusion-exclusion
side of) your formula for E P ( ⋃ n
i=1 A i)?
(d) How have you proved your conjecture for the picture enumerator of
the union of the sets A i ?
(e) How can you get the formula for the principle of inclusion and exclusion
from your formula for the picture enumerator of the union?
Problem 234. Frequently when we apply the principle of inclusion and
exclusion, we will have a situation like that of part (d) of Problem 231.d.
That is, we will have a set A and subsets A 1 , A 2 ,...,A n and we will want
the size or the probability of the set of elements in A that are not in the
union. This set is known as the complement of the union of the A i sinA,
and is denoted by A \ ⋃ n
i=1 A i,orifA is clear from context, by ⋃ n
i=1 A i. Give
the fomula for ⋃ n
i=1 A i. The principle of inclusion and exclusion generall
refers to both this formula and the one for the union.
We can find a very elegant way of writing the formula in Problem 234 if we
let ⋂ i:i∈∅ A i = A. for this reason, if we have a family of subsets A i of a set A, we
define1 ⋂ i:i∈∅ A i = A.
1For those interested in logic and set theory, given a family of subsets A i of a set A, we define ⋂ i:i∈S A i
to be the set of all members x of A that are in A i for all i ∈ S. (Note that this allows x to be in some other
A j s as well.) Then if S = ∅, our intersection consists of all members x of A that satisfy the statement “if
i ∈∅, then x ∈ A i .” But since the hypothesis of the “if-then” statement is false, the statement itself is
true for all x ∈ A. Therefor ⋂ i:i∈∅ A i = A.