Unformatted text preview: Â± 2 n n Â² = Â± n Â²Â± n n Â² + Â± n 1 Â²Â± n n1 Â² + Â± n 2 Â²Â± n n2 Â² + Â·Â·Â· + Â± n n Â²Â± n Â² Note that this is a special case of Vandermondeâ€™s 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 âˆª T . We can also count this same quantity as follows: For each k between 0 and n , there are ( n k )( n nk ) ways to choose k elements from S and the remaining nk elements from T . Thus, the total number of ways is also X Â± n k Â²Â± n nk Â² which is exactly the RHS....
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 Summer '08
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 Math, Natural number, Playing card, standard 52card deck, Rob Bayer

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