Proof C(n,r) = C(n, n-r)
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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$\begingroup$$ 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.
$\endgroup$ $\begingroup$How to show this? Choosing $r$ sets out of n possibilities is the same as selecting $n-r$ sets not to choose.
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