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Unformatted text preview: Considering the right hand side of this expression, we have n−1 r−1 + n−1 r (n − 1)! (n − 1)! + (n − 1 − r + 1)!(r − 1)! (n − 1 − r )!r ! (n − 1)! (n − 1)! = + (n − r )!(r − 1)! (n − 1 − r )!r ! n! r n−r = + (n − r )!r ! n n n = , r = and the result is proven. Problem 8 (selecting r people from from n men and m women) We desire to prove n+m r = n 0 m r + n 1 m r−1 + ...+ n r m 0 . We can do this in a combinatorial way by considering subgroups of size r from a group of n men and m women. The left hand side of the above represents one way of obtaining this identity. Another way to count the number of subsets of size r is to consider the number of possible groups can be found by considering a subproblem of how many men chosen to be included in the subset of size r . This number can range from zero men to r men. When we have a subset of size r with zero men we must have all women. This can be done in n m ways. If we select one man and r − 1 women the number of subsets that meet 0 r n m this criterion is given by . Continuing this logic for all possible subset of 1 r−1 the men we have the right hand side of the above expression. Problem 9 (selecting n from 2n) From problem 8 we have that when m = n and r = n that 2n n Using the fact that = n k n 0 = 2n n which is the desired result. n n + n 1 n n−1 + ...+ n n n 0 . n n−k = n 0
2 the above is becomes + n 1
2 + ...+ n n 2 , ...
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This note was uploaded on 02/25/2011 for the course STAT 418 taught by Professor G.jogeshbabu during the Winter '08 term at Pennsylvania State University, University Park.
- Winter '08