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Unformatted text preview: This is often written 5
, read “5 choose 3.” More generally,
3
n
k = n!
.
k !(n − k )! An example. How many ways can one choose a committee of 3 out of 10
10
people? Answer:
.
3
An example. Suppose there are 8 men and 8 women. How many ways
can we choose a committee that has 2 men and 2 women? Answer: We
8
8
can choose 2 men in
ways and 2 women in
ways. The number of
2
2
88
.
committees is then the product:
22
Let us look at a few more complicated applications of combinations. One
example is the binomial theorem:
n
n (x + y ) =
k=0 n k n−k
xy .
k To see this, the left hand side is
(x + y )(x + y ) · · · (x + y ).
This will be the sum of 2n terms, and each term will have n factors. How
many terms have k x’s and n − k y ’s? This is the same as asking in a sequence
of n positions, how many ways can one choose k of them in which to put x’s?
n
n
The answer is
, so the coeﬃcient of xk y n−k should be
.
k
k
One can derive some equalities involving these binomial coeﬃcients by
combinatorics. For example, let us argue that
10
4 = 9
9
+
3
4 without doing any algebra. Suppose we have 10 people, one of whom we
10
decide is special, denoted A.
represents the number of committees
4
12 ...
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This note was uploaded on 12/29/2011 for the course MATH 316 taught by Professor Ansan during the Spring '10 term at SUNY Stony Brook.
 Spring '10
 ansan

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