Assignment 1 Soln

Assignment 1 Soln - MATH 239 Assignment 1 Solutions This...

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Unformatted text preview: MATH 239 Assignment 1 Solutions This assignment is due at noon on Friday, May 15, 2009, in the drop boxes opposite the Tutorial Centre, MC 4067. 1. Let integers 0 ` k n be given. Consider the binomial identity n k k ` = n ` n- ` k- ` . (a) Give an algebraic proof of this identity. Solution: Using the formula ( n k ) = n ! ( n- k )! k ! we find LHS = n ! ( n- k )! k ! k ! ( k- ` )! ` ! = n ! ( n- k )!( k- ` )! ` ! and RHS = n ! ( n- ` )! ` ! ( n- ` )! ( n- k )!( k- ` )! = n ! ( n- k )!( k- ` )! ` ! , so LHS=RHS as required. (b) Give a combinatorial proof of this identity. Solution: We define a set S of pairs of subsets as follows: S = { ( A, B ) : A B { 1 , 2 , . . . n } , | A | = `, | B | = k } . We calculate the size of the set S in two different ways. First, we choose k elements of { 1 , 2 , . . . , n } in all possible ways to form the set B , then from the k elements of B we choose ` elements in all possible ways to form the set A . Therefore the number of elements of S is ( n k )( k ` ) . Now we calculate the size of S by choosing the set A first. We can do this by choosing ` elements of { 1 , 2 , . . . , n } in all possible ways. Then to form the set B , we must choose k- ` elements from the remaining n- ` elements of { 1 , 2 , . . . , n } that were not chosen to be elements of A . There are therefore ( n- ` k- ` ) such choices. Therefore the size of S is ( n ` )( n- ` k- ` ) . Thus we have ( n k )( k ` ) = | S | = ( n ` )( n- ` k- ` ) as required....
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Assignment 1 Soln - MATH 239 Assignment 1 Solutions This...

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