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PROFESSOR: The binomial
theorem extends

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to a thing called the
multinomial theorem,

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whereas instead of taking a
product of a sum of two things,

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you'd take the product
of a sum of k things

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to get the multinomial theorem.

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And what underlies it
is a rule that we're

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going to call the bookkeeper
rule, and here's why.

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So, the bookkeeper rule
is about the question of,

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look at the word bookkeeper
and ask how many different ways

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are there to scramble the
letters in this word that

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actually are distinguishable?

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The point being that the two
o's are indistinguishable,

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so the order in which they
appear doesn't matter.

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Likewise, the three
e's and the two k's.

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Well, how do we
answer this question?

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The simple way to
do it to begin with

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is to label all of
the indistinguishable

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letters with subscripts to
make them distinguishable.

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So, I'm going to put subscripts
1 and 2 on the o's, 1 and 2

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on the k's, and 1,
2, and 3 on the e's.

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Now, all the 10 letters
are distinguishable.

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And if I ask how
many ways are there

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to permute these 10
letters, the answer,

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we know by the
generalized product

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rule is simply 10 factorial.

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Now, my strategy is going
to be to use the division

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rule to count the number
of patterns of the letters

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in the word with no subscripts.

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And the way I'm
going to do that is

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take one of these subscripted
words and erase the subscripts.

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So, I'm going to map it to the
same permutation of letters

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with no subscripts.

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I've just done that.

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Here I've taken an
arbitrary permutation

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of the subscripted word, and
then I've erased the subscripts

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and consolidated the letters.

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And I wind up with
this permutation.

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Now, if I want to
count the number

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of unsubscripted
permutations, then I simply

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figure out that this
mapping is K to 1,

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and I'm going to then divide by
K. Well, how many to 1 is it?

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Well, how many subscripted
words map to this given pattern?

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The answer is the subscripts
on the o's don't matter,

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so there's two possible orders
in which those subscripts might

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appear.

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Subscripts on the
k's don't matter.

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There's two possible
orders in which

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those subscripts might appear.

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Subscripts on the
e's don't matter.

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Three possible orders, or 3
factorial possible orders that

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the subscripts might
occur in the e's.

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The net result is that with
two o's, two k's, and three

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e's, the mapping is 2
factorial by 2 factorial

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by 3 factorial to 1.

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And that instantly gives
us, by the division rule,

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that the total number
of permutations

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of the letters in
the word bookkeeper

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is 10 factorial over 2 factorial
times 2 factorial times 3

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factorial.

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More generally by
the same reasoning

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if, I look at a sequence of
n letters, of which n1 are

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a's and n2 are b's up
through nk are z's, then

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the number of permutations
of those letters

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with the repeated
a's, b's, and Z's is

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n factorial divided
by n1 factorial times

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n2 factorial through
nk factorial.

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And this formula occurs so
often that it has a name.

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It's called a
multinomial coeff--

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there's a name for it
written in this format,

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n over n1, n2 through nk.

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You could start to say n
choose n1 choose n2 choose nk,

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if you're thinking
about how we pronounce

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the binomial coefficients.

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The convention is that the
sum of the ni's is supposed

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to be equal to the numerator n.

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This is called a
multinomial coefficient.

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So, n factorial divided by
this product of factorials

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is written in somewhat
shorter notation

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without the factorials as
a multinomial coefficient.

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Binomial coefficient, by
the way, are a special case.

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When we write n choose
k, if we wrote it

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as a multinomial
coefficient, you'd

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have to write it as n choose
k and then choose n minus k.

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So, we can apply this to think
about words and coefficients

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and expanding things that
are more than binomials.

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So, let's look at expanding
a quintomial, a sum of five

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things, E, M, S, T,
and Y. And I raise that

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to the seventh power.

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So, that means in these products
of seven of these terms,

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I'm looking at words
of length seven

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whose components are the
letters E, M, S, T, and Y.

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And So, if I multiply this out,
applying the distributive law,

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I would wind up with
5 to the 7th terms,

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each of them consisting of
a permutation of the letters

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E, M, S, T, and Y.

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And if I ask what's the
coefficient in that expansion

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of the term E, M,
S cubed, T, Y, it's

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exactly the number of ways
of permuting these five

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letters, a word of length seven
made out of these five letters

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with three occurrences
of S. In other words,

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the coefficient of E, M, S
cubed, T, Y in this product

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is the number of ways of
rearranging the letters

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in this sequence of seven.

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It's the word
systems, which is why

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we chose it to be rememberable.

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How many ways are
there to rearrange

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the letters in the word
systems by the bookkeeper rule?

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There are seven.

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Choose 1, 1, 3, 1, 1.

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Let's do another example.

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What's the coefficient
of BA cubed

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N squared if I expand
this trinomial, B plus A

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plus N to the sixth power?

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Well, now again I have
3 to the 6th terms.

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How many of them involve
a B, three A's, and two

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N's by the bookkeeper rule?

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It's the number of
ways-- well, it's

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the number of ways
of rearranging

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the letters in the word banana.

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And by the bookkeeper
rule, that's

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six with subscripts 1, 3, and 2.

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More generally, this is what
the multinomial theorem says.

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If I look at the
coefficient of the term--

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a product of Xi to the
ri's in an expansion

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of a k-nomial, a sum of k
distinct variables raised

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to the n-th power, now
I've got if I expanded this

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out using the distributive
law without collecting terms,

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I'd have k to the n
terms, each of which

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was a permutation of the X1's
through Xk's, with repeats.

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And then if I ask, how
many of those products,

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if any of these k variables
have this many X1's, this many

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X2's, through this many rk's--
this many Xk's, I'm asking

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again a bookkeeper question.

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And the answer is n
choose r1, r2 through rk.

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So, now we're ready
for the record

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to state the general
multinomial formula.

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If I take a sum of k terms,
a k-nomial to the nth power,

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then expressing it
in concise notation,

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it's the sum over r1 through rk
summing to n of the multinomial

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coefficient n r1 through rk
times this product of Xi's.

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I'm not putting a
highlighted box around it,

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because this is not a
formula which is particularly

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important to memorize.

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And it's clearly all
clogged up with subscripts.

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But nevertheless, it's good to
have sometimes for the record.

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And next week, we will
continue with this theme

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about the connection between
counting and algebra.

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And in particular, not
only ordinary polynomials

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as we've been looking at
here with a product of sums,

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but in fact, infinite
polynomials or infinite series

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when we pick up generating
functions next week.