WEBVTT
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We now come to our last major
class of counting problems.
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We will count the number of ways
that a given set can be
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partitioned into pieces
of given sizes.
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We start with a set
that consists
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of n different elements.
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And we have r persons.
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We want to give n1 items to
the first person, give n2
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items to the second
person, and so on.
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And finally, we want
to give n-sub-r
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items to the rth person.
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These numbers, n1, n2, up to nr
are given to us, how many
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items each person should get.
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And these numbers must add to
n so that every item in the
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original set is given
to some person.
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We want to count to the number
of ways that this can be done.
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This is the number of ways that
we can partition a given
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set into subsets of
prescribed sizes.
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Let's use c to denote
the number of
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ways this can be done.
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We want to calculate
this number c.
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Instead of calculating directly,
we're going to use
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the same trick that we employed
when we counted
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combinations and derived the
binomial coefficient.
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That is, we're going to
consider, in a much simpler
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counting problem, the problem
of ordering n items, taking
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the n items in our original
set and putting them in an
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ordered list.
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Of course, we know in how many
ways this can be done.
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Ordering n items can be done
in n factorial ways.
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This is the count
of the number of
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permutations of n items.
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But now let us think of a
different way of ordering the
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n items, an indirect way.
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It proceeds according to
the following stages.
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We start with the n items.
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And we first distribute them
to the different persons.
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Having done that, then we ask
person one to take their
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items, order them, and put
them in the first n1
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slots of our list.
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Then person two takes their
items and puts them into the
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next n2 slots in our list.
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We continue this way.
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And finally, the last person
takes the items that they
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possess and puts them
in the last n-sub-r
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slots in this list.
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In how many ways can this
process be carried out?
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We have c choices on how
to partition the
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given set into subsets.
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Then person one has n1 factorial
choices on how to
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order the n1 items that
that person processes.
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Person two has n2 factorial
choices for how to order the
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n2 items that it possesses,
and so on until the last
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person, who has nr factorial
choices for
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ordering their elements.
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This multi-stage process results
in an ordered list of
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the n terms.
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This is the number of ways these
multi-stage process can
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be carried out.
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On the other hand, we know that
the number of possible
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orderings of the items
is n factorial.
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So we have this equality.
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We can solve this for c.
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And we find the answer, that the
number of ways that the n
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items can be partitioned into
subsets of the given sizes is
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n factorial divided by the
product of the factorials of
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the different ni's.
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This particular expression
is called the multinomial
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coefficient, and it generalizes
the binomial
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coefficient.
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The binomial coefficient was
referring to the case where we
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essentially split our set into
one subset with k elements,
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and then the second subset gets
the remaining elements.
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So the special case where r is
equal to 2, and n1 is equal to
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k, n2 equals to n minus k,
this corresponds to a
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partition of a set into two
subsets, or what is the same
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just selecting the first subset
and putting everything
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else in the second subset.
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And you can check that in this
particular case, the
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expression for the multinomial
coefficient agrees with the
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expression that we had derived
for the binomial coefficient.