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PROFESSOR: The
arguments that we just

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reviewed for proving
inclusion-exclusion

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by looking at how many
times points are counted

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can be made perfectly rigorous.

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In fact, we'll do that
in a later segment,

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but it's interesting
and good practice

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to realize that they
can also be proved just

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from some simple set
theoretic identities using

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the ordinary disjoint sum rule.

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So how am I going to prove the
inclusion-exclusion principle

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for two sets?

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The size of A union
B is the size of A

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plus the size of B minus
the size of A intersect B,

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and the idea is just break up
A union B into disjoint sets

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because once they're disjoint
sets, I can add up their sizes.

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So if we look at A
union B, A union B

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can be expressed as the
union of two disjoint sets,

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namely A-- the round
blue circle-- and what's

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left-- the points in B that
are not in A, so this lighter

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orange-colored region.

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So A and B union and B minus
A-- so these are the points in A

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and these are the points that
are in B that are not in A

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are two disjoint sets
whose union is A union B.

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That means that I
know the size of this.

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It's just the size of A
plus the size of B minus A

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because they're disjoint, so
we conclude by the sum rule

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that the size of A union B
is equal to the size of A

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plus the size of B minus A.

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So now I need a
little lemma that

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says that the size of B
minus A is the size of B

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minus the size of
A intersection B.

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If I get that, then I've
proved inclusion-exclusion

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because now I have A plus
B minus A intersection B.

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So we need a lemma.

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The lemma says that
the size of B minus A

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is equal to the size of B minus
the size of A intersect B.

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So here we're back
to the Venn diagram.

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A is blue, B is reddish,
and the intersection region,

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that lens-shaped region,
is shown in purple,

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and we want to prove this lemma.

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Well, again, what we can do
is look at the set B broken up

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into two pieces.

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B can be expressed as a
disjoint union of the part of B

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that's in A union, the part
of B that's not in A. That is,

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for any set B and
any set A, B is

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equal to the B points
that are in the A

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and the B points that are not
in A covers all the cases.

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And this, again, is
a disjoint union.

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So we conclude immediately from
the sum rule that the size of B

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is equal to the size
of B intersection

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A plus the size of B minus
A. And then just transposing

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this term for the
size of B minus A

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to the left-hand side of the
equality, I've proven the lemma

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and we're done.

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Now, inclusion-exclusion
for three sets,

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we've said before it's this
slightly more complicated thing

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where you've got a sum of
the intersections of one

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set minus the sizes of the
intersections of two sets

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plus the size of the
intersection of three sets.

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And that generalizes to the
following somewhat messy

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formula, but let's look
at it closely together.

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If I want to know what's
the size of A1 through An--

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if I have n sets potentially
overlapping A1 A2 through An

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and I want their union, and I
can express the union of them

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in terms of a sum of
sizes of intersections,

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and here's the formula.

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Let's read this slowly together.

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So this is a sum over every
possible subset of the indices

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1 through n that's not empty.

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So this sum is ranging over S,
and it can do it in any order.

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It's typical to write the sum
where you sum up first all

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the sets S of size 1 and then
sum up all the sets S of size 2

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and sum-- but that's
not necessary.

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We just sum in any order for
every subset of the indices,

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the size of the intersection
of the Ai's that

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are in this set of
indices specified by S.

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So this is just the intersection
of those A's that s specifies.

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Now, what's the sign of
that size of intersection?

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As we said, if it's of odd size,
I want it to count positively.

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So if I take minus 1
to the odd size plus 1,

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I get an even power of minus
1, so it comes out to be 1.

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If, on the other
hand, the size of S

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is even-- so I'm
taking an intersection

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of an even number of sets--
then this number to the plus 1

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is odd.

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I'm taking minus
1 to an odd power,

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and sure enough, I'm
getting the negative sign

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on all the intersections
of odd size.

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So that's what this rather
concise but hairy formula

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looks like.

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Here we have an intersection
over the Ai's where

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the i is specified by
the set S of indices,

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and I sum up these terms over
every possible nonempty set

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S. That is the generalized
form of inclusion-exclusion

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for n sets.

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Now, how do we prove this?

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Well, there's lots
of ways to prove it.

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The simplest way is to do
it actually by induction.

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It's not very hard
to do by induction.

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You just use the two-set version
of inclusion-exclusion, which

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we've proved rigorously
using the disjoint sum rule,

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and you go from the union of
the first n sets plus the nth

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and apply the formula and
simplify, and that works fine.

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The other problem with it is
[? it's ?] not really very

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informative.

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You prove the theorem all right.

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As is frequently the case
with induction proofs,

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you have the right
induction hypothesis,

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the proof is kind of
mechanical, but you don't really

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learn much from it-- not
always, but in this case,

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I don't think you do.

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A second way to do it is by
using the binomial theorem

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and counting.

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This is a way to make rigorous
the argument that said,

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OK, we counted two points in
the intersection of two things

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twice, so we had to
subtract them away,

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and then that meant
that we were not

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counting the things that were
in the intersection of three

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not at all, and so on.

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And we've talked through
that argument informally,

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and I will do that in a
next segment of making

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that argument a little
bit more precise to prove

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the general theorem.

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And--

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[AUDIO OUT]

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Law from algebra, we can just
look at a product of sums

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and understand how it expands
into a sum of products

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by the distributivity law.

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And that's worked out in a
problem that is in the text,

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and I'm not going to
do that in a video.