Hello just want to see if my proof is right, and if not could someone please guide me because I am not clearly seeing the steps to this proof. I don't know if I correctly solve the proof in the second to last step. If I did any mistake it would be great if someone could point at it. $$ C(n, n-r) = \frac{n!}{r!(n-r)!} = \frac{n!}{(n-r)!(n-(n-r))!} = \frac{n!}{(n-r)! (r)!} = C(n,r) $$
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I've edited it, the proof looks fine to me. Can you think of a "combinatorial" proof of this equality? – Sarvesh Ravichandran Iyer Mar 27 '17 at 02:38
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The second formula after $C(n,n-r)$ should be the first formula. The third should be the second. And, the first should be the third. – J126 Mar 27 '17 at 02:39
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proof could be just that whenever you select $r$ from $n$ you are also selecting $n-r$ from $n$. – miniparser Mar 27 '17 at 02:54
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Thank you for the feedback. – Alan Mar 27 '17 at 03:23
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Let c(n,r) be the set of r element subsets taken from some n element set and define c(n,n-r) similarly. The complement map (restricted) bijects these two sets and so they have the same cardinality. – Jacob Wakem Mar 27 '17 at 05:39
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$ C(n, k)$ denotes the number of ways to select $k$ out $n$ objects without regard for the order in which they are selected. To prove $C(n,r) = C(n, n-r)$ one needs to observe that whenever $k$ items are selected, $n-k$ items are left over, (un)selected of sorts.
User8976
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How to show this? Choosing $r$ sets out of n possibilities is the same as selecting $n-r$ sets not to choose.
LM2357
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Jacob Wakem
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