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| + | 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? | ||
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| + | --[[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]] | ||
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| + | For combinatorial proof, are we to think about the same subject in two different ways?? | ||
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| + | Once I manipulate the formula, I get the other side of equation but.. maybe I do not understand the true meaning of the equation.. like how it works and stuff.. | ||
| + | --[[User:Kangw|Kangw]] 00:35, 5 February 2009 (UTC) | ||
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| + | I think that you have the right idea Kangw. You did work on both sides of the equation and were able to get them to a similar point. I think that this is the right idea. --[[User:Elhalsme|Elhalsme]] | ||
Latest revision as of 06:23, 5 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
For combinatorial proof, are we to think about the same subject in two different ways??
Once I manipulate the formula, I get the other side of equation but.. maybe I do not understand the true meaning of the equation.. like how it works and stuff.. --Kangw 00:35, 5 February 2009 (UTC)
I think that you have the right idea Kangw. You did work on both sides of the equation and were able to get them to a similar point. I think that this is the right idea. --Elhalsme
