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Unformatted text preview: 2 n n = n n n + n 1 n n-1 + n 2 n n-2 + + n n n Note that this is a special case of Vandermondes Identity, and you may only use that theorem if you prove it here. Consider two sets, S and T that are disjoint and each have n elements in them. Then the LHS counts the number of ways to choose n elements from S T . We can also count this same quantity as follows: For each k between 0 and n , there are ( n k )( n n-k ) ways to choose k elements from S and the remaining n-k elements from T . Thus, the total number of ways is also X n k n n-k which is exactly the RHS....
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This note was uploaded on 08/31/2009 for the course MATH 55 taught by Professor Strain during the Summer '08 term at University of California, Berkeley.
- Summer '08