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Unformatted text preview: doesn’t matter, we will have the letters a,b,c 6 times, because there are 3! ways of arranging 3 letters. The same is true for any choice of three letters. So we should have 5 × 4 × 3 / 3!. We can rewrite this as 5 · 4 · 3 3! = 5! 3!2! This is often written 5 3 , read “5 choose 3 .” More generally, n k = n ! k !( n k )! . Example. How many ways can one choose a committee of 3 out of 10 people? Answer. 10 3 . 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 can choose 2 men in 8 2 ways and 2 women in 8 2 ways. The number of committees is then the product: 8 2 8 2 . Let us look at a few more complicated applications of combinations. One example is the binomial theorem: ( x + y ) n = n X k =0 n k x k y n k . To see this, the left hand side is ( x + y )( x + y ) ··· ( x + y ) . This will be the sum of 2 n 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’s?...
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This note was uploaded on 12/29/2011 for the course MATH 317 taught by Professor Wen during the Spring '09 term at SUNY Stony Brook.
 Spring '09
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