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WetWired

WetWired said: [Goto]

Wait! What an obtuse way to say that n choose k = n choose k or n choose n-k!

Why not just $\reverse {n\choose k} = {n\choose n\cdot{(-1)^r+1\over2} + k\cdot(-1)^{r+1}},\,r \in \mathbb{Z}$ or even $\reverse {n\choose k} = {n\choose n-k}$

Yes, my "formula" is just the result of using the identity $\reverse {n\choose k} = {n\choose n-k}$ repeatedly, followed by a little algebra:

$\reverse {n\choose k} = {n\choose n-k} = {n\choose n-(n-k)} = {n\choose n-(n-(n-k))} = \dots$

I just thought the symmetry between the sum and product (and their upper limits) was fun.