WEBVTT
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Hello.
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Let us see the first exercise
of the binomial coefficient
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in practice step,
always into week 3.
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We want to count
in how many ways
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one can choose 3 cards
in a deck of 10 cards.
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Ok, when you have to choose
without any particular order,
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elements in a set of
different elements,
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you have to use the
binomial coefficients.
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In particular,
for this exercise,
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we have that the number of
ways in which we can choose
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3 cards in a deck of
10 cards is exactly
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equal to the binomial 10 over 3.
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This number-- now we'll
see what is this number.
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But this symbol is
exactly the number
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of possible choices of 3
elements among 10 elements.
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This is by definition equal
to the factorial of 10
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over the factorial of 10 minus
3 times the factorial of 3.
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Now, we can simplify
the factorial of 10
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and the factorial of 7, and
what remains is exactly 10 times
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9 times 8.
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And at the denominator, we
have only the factorial of 3,
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which is 3 times 2, if you want,
times 1, but this, of course,
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is not necessary.
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And now what do we get?
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You easily can simplify
this fraction, 3 and 9,
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and you get 3 here and 2
and 8, and you get 4 here.
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And what you get?
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10 times 3, which are 30
times 4, which is 120.
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Then we can choose 3
cards in a deck of 10
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cards in 120 different ways.
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Let us see now the
second exercise
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of the binomial coefficients
in practice step.
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We have to prove that the
binomial coefficient, n over k,
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is equal to the binomial
coefficient n over n minus k
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for any k and n, which
satisfy these equalities.
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Then let us start to with first
just considering the definition
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of the binomial coefficient.
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You know that n over k
is equal by definition
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to the factorial of n over
the factorial of n minus k
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times the factorial of k.
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And what is the definition
of this other binomial
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coefficient?
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You have that n over n minus k
is equal to the factorial of n.
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And in the denominator, we have
the factorial of the difference
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between n and n minus k--
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n minus n minus k factorial--
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times the factorial
of n minus k.
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But this is exactly n factorial.
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And here what you have,
n minus n minus k,
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and you remain with k factorial
times n minus k factorial.
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And this and this
are perfectly equal.
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But this is just an
algebraic proof, if you want.
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But it's much more interesting
to consider a proof which
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regards the meaning of
the binomial coefficient
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in a combinatorial sense.
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Remember, what means n over k?
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This number counts
in how many ways
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you can choose k elements
among n different elements.
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That is, you have n
different elements,
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and you want to choose k
elements among these n.
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For example, you can decide
to choose these k elements
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among these n elements.
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In general, in how many ways can
you choose k elements among n,
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binomial coefficient
n over k times.
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But observe.
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What means to choose these
k elements among these n
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elements is exactly like
to decide to discard
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these n minus k elements.
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Just no.
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You put away these
n minus k element,
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and you remain with
your k elements.
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Therefore, to choose
k elements among n
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is equivalent to
discard n minus k
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elements among your n elements.
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Then to choose k
elements among n,
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and you can do the same
exactly binomial coefficient
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n over k times--
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is equal to choose
the n minus k element
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to discard among
your n elements.
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And therefore, these two
binomial coefficients
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are equal.
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Ciao.
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