elemprob-fall2010-page12

# elemprob-fall2010-page12 - This is often written 5 , read...

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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.

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