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Unformatted text preview: 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 n positions, how many ways can one choose k of them in which to put x ’s? The answer is n k , so the coefficient of x k y n k should be n k . One can derive some equalities involving these binomial coefficients by combinatorics. For example, let us argue that 10 4 = 9 3 + 9 4 without doing any algebra. Suppose we have 10 people, one of whom we decide is special, denoted A ....
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 Fall '06
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