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PROFESSOR: There are
two generalizations

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of the bijection rule
and the product rule that

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come up all the time and
play an essential role

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in the repertoire
of any counter.

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So let's look at those.

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The first of these
is a generalization

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of the product rule.

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And let's see an instance
where it comes up.

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Suppose I wanted to count
the number of lineups

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of five students in the class.

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So if I let S be the number
of students, and let's say,

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for the afternoon
session S is 91, then

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the number of lineups of
five students-- if I use

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the ordinary product
rule, I would

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get-- I'm talking about
S to the fifth, that is,

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sequences of length
five of elements of S.

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And so the product
rule would say,

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take 91 to the
fifth as the number

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of lineups of five students.

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And that would be correct if the
same student could appear twice

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in line, but that,
of course, isn't

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possible with real students.

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So the lineups have no repeats.

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And what we're really
counting is the number

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of those sequences of
length five of students

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with no repeats.

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And the generalized
product rule tells

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you quite straightforwardly
how to count those.

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Namely, there are 91 ways
to choose the first student

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among the 91.

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And whichever first
student you've chosen,

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that leaves 90 other students
you could choose to be second.

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And once you've chosen
the first two, that

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leaves 89 students you
could choose for the third,

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and 88 for the fourth,
and 87 for the fifth.

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And the formula then is 91
times 90 times 89, 88, 87,

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for the number of sequences
of distinct students

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of length five.

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Now, one nice way to
express the 91 down

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to 87, in terms of factorials,
is its 91 factorial,

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which is the product
from 1 to 91,

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and divided by the
product from 1 to 86,

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which cancels out the first
86 terms in 91 factorial,

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leaving me with exactly
87 through 91 product.

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So the second rule is a sort
of obvious generalization

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of the bijectional, but I'm
getting ahead of myself.

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Let's state the generalized
product rule in general.

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So if we let Q be a set
of length-k sequences

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with the following
property, there

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are n1 possible first elements
among these length-k sequences.

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And for every one of the
first possible elements,

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if you look at the
number of tuples

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with what the second possible
coordinates for a given

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first coordinate,
it's always n2.

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And likewise, if you
look at the number

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of possible third coordinates
given the first two, it's n3

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and it's uniform no matter
what the first two are.

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Then if you have this
kind of a set up,

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which is exactly what happens
when you're picking one student

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after another and
they can't compete,

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you discover that the
length-k sequences with n1 ,

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first possible choices, n2,
second possible choices,

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down through nk, k-th possible
choices is n1 through nk.

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So that's the statement
of the generalized product

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rule in the magenta box.

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Now, we come to the generalized
bijection rule, which

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is called the division rule.

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And a simple, memorable
way to illustrate

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is if you wanted to count the
number of students in class

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6.042, you could count the
number of students fingers

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and divide by 10.

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Now, it's probably harder to
count fingers than students,

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so this is not meant
as a practical method.

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But it illustrates a basic
and straightforward idea.

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Of course, it's
implicitly assuming

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that we don't have any
instances of amputations

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or polydactylism, and that,
in fact, every student

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has exactly 10 fingers.

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OK, so in general,
the division rule

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can be stated this
way, if I have

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a total function from a set A to
a set B, domain A, co-domain B,

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and this mapping is k-to-1,
then the cardinality of A

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is simply k times the
cardinality of B. So k-to-1

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means that exactly k A
elements hit each B element.

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Another way to say it is that
there are exactly k arrows

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into every element of B. So
then the number of arrows

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is simply k times B. And if
you have a total function on A,

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the number of arrows is
equal to the size of A,

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and that's where
we get the formula.

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OK.

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And that's the generalized
bijection rule.

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Let's apply it in
a crucial example

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that is absolutely basic and
we'll be using repeatedly.

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Suppose that I want to know how
many possible subsets of size

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four are there from the
numbers 1 through 13?

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So I have 13 possible
numbers that I can choose.

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I want to pick out
any four of them

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and I want to know how many
ways are there to do that.

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And we'll do that by finding
a mapping from things

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we know how to count to
these particular subsets.

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So what we know how
to count is if I

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let A be the set of all
permutations of 1 through 13,

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then I know that the size of A
is 13 factorial because there's

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13 choices for the first
element of the permutation, 12

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for the second, down
to one for the 13th.

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And let's let B be this
object that I want to count,

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namely, the set of size four
subsets of 1 through 13.

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And I want to find
a mapping from A

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that I know how to count
to B that I don't yet

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know how to count,
but in a way where

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I can figure out that
it's k-to-1 for a k

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that I can also count.

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How do I do that?

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Well, let's take an
arbitrary permutation of A,

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that is to say, a
sequence of the elements

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of A in some order-- call
them a1, a2, through a13.

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So these numbers a1 through a13
are those numbers 1 through 13

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in some unknown order.

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And I'm going to map
a permutation of A,

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like this, to its
first four elements.

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Just take the first four
elements of the permutation

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and map them to
the set consisting

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of those four elements.

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Now, since this
is a permutation,

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these elements
are all different,

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so I really do get a set of
four different things here,

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a1, a2, and a3.

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And a4 is supposed
to be different.

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This gives me a very
well-defined total function

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from a permutation of 13
numbers to set of its first four

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elements.

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And now what we want to know is
what kind of a mapping is this?

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And I'm going to argue that
it's k-to-1 for a k that's

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not very hard to count.

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So when I look at
what other things map

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to the set a1, a2, a3, a4,
we mapped a permutation

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to its first four elements.

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And if we've got a1 through a4
as the set, what other things

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map to that set a1, a2, a3, a4?

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Well, the answer
is any permutation

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with the same first four
elements, but possibly

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in a different order,
because we're just

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going to take the
first four in sequence

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and map them to the set
of those first four.

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The order in which the
first four doesn't matter.

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OK?

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And likewise, the order of
the remaining nine elements,

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5 through 13, also
doesn't matter.

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Whatever they are, if I have
a given set of four elements

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to start, no matter what
the remaining 9 are,

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they're going to map to the
same subset of four elements.

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So there are 4
factorial possible ways

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that the first four
elements can be permuted.

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And there are 9 factorial
ways that the last nine

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elements can be permuted.

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And every one of these goes to
the same set of four elements,

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a1 through a4.

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And those are the only
ones that go there.

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And so what we've figured
out is that the mapping

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of these kind of sequences with
the given four elements first

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in some order and
the remaining nine

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elements in some other
order is 4 factorial times 9

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factorial-to-1.

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There are 4 factorial times
9 factorial permutations

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that map to any given
set of four elements.

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And that means that by
applying the division rule,

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I've discovered that the
size of A, which I know

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is 13 factorial,
is equal to that k

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of the k-to-1 of 4 factorial
times 9 factorial times

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the size of B.

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B is the subsets of size four
that I'm trying to count.

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And so what I get is
that the size of B

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is simply 13 factorial
divided by that k, 4

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factorial 9 factorial, 13
factorial over 4 factorial

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9 factorial.

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And this number
comes up so often

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that it has this
special notation called

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binomial coefficient notation,
which we read as 13 choose 4.

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In general, if I
have an n element set

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and I'm going to choose a subset
of m of them-- generalizing

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this argument because
the 4 and the 9

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and the 13 were
completely arbitrary

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and the argument
works in general--

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is that the number of ways
to choose a set of m elements

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among n is n choose m.

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And the definition
of n choose m is

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n factorial over
the m factorial ways

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to permute the first m elements
and the n minus m factorial

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ways to permute the
remaining n minus m elements.

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And again, that notation,
the binomial coefficient,

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is called n over
m is n choose m.

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This is an absolutely
fundamental formula

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that you need to remember
because we will be using it

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constantly and repeatedly.