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--[[User:Jberlako|Jberlako]] 16:38, 4 February 2009 (UTC) | --[[User:Jberlako|Jberlako]] 16:38, 4 February 2009 (UTC) | ||
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| + | I think we're talking about the same problem -- for the left side, you can simply consider it choosing 2 elements from 2n elements. For the right side -- imagine that you divided the 2n elements into 2 sets of n. You first account for the possibility that you pick both elements from the same half, thus the nC2 (which can happen for both halves, thus you multiply by 2). Then you add the number of ways to pick one element from each set -- yielding n * n = n^2. The result is that you are picking two items from 2n elements in both cases. | ||
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| + | --[[User:Kshen|Kshen]] | ||
Revision as of 13:10, 4 February 2009
Has anyone managed to do the combinatorial proof for this one? I am completely stuck. I have the vague idea that it is related to the binomial theorem, but all of my reasoning stalls after expressing (x+y)^2n = (x+y)^n*(x+y)^n. I could be going in a dead end, of course. Any suggestions?
--Jberlako 16:38, 4 February 2009 (UTC)
I think we're talking about the same problem -- for the left side, you can simply consider it choosing 2 elements from 2n elements. For the right side -- imagine that you divided the 2n elements into 2 sets of n. You first account for the possibility that you pick both elements from the same half, thus the nC2 (which can happen for both halves, thus you multiply by 2). Then you add the number of ways to pick one element from each set -- yielding n * n = n^2. The result is that you are picking two items from 2n elements in both cases.
--Kshen
