Solutions from a student (dragged) 18

# Solutions from a student (dragged) 18 - Considering the...

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Considering the right hand side of this expression, we have n - 1 r - 1 + n - 1 r = ( n - 1)! ( n - 1 - r + 1)!( r - 1)! + ( n - 1)! ( n - 1 - r )! r ! = ( n - 1)! ( n - r )!( r - 1)! + ( n - 1)! ( n - 1 - r )! r ! = n ! ( n - r )! r ! r n + n - r 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
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