This preview shows page 1. Sign up to view the full content.
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 , ...
View
Full
Document
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
 G.JOGESHBABU
 Probability

Click to edit the document details