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The task is to find the parity of ${2n\choose 2k+1}$ where $n,k\in\mathbb{N}$. How can I do that?

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Could you be more specific about what it is you seek to prove? "The parity of $\binom{2n}{2k+1}$" is not a claim that can be true or false, so it is not a priori something it makes sense to prove. – Henning Makholm Mar 3 '12 at 22:55

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Hint $\rm\displaystyle\ \ k {n\choose k} =\ n {n-1 \choose k-1 }\:$ so $\rm\:k\:$ odd, $\rm\:n\:$ even $\:\displaystyle\rm\Rightarrow {n \choose k}\:$ is $\:\ldots$

For the parity of the general case see here.

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Yes, but I didn't have congruences in school yet, anyway this case is easy, and don't need them, thanks :-) – xan Mar 3 '12 at 23:51
@xan Ah, I only now noticed that was you too in the prior question. The other cases in my prior answer can be handled similarly as above, i.e. by clearing fractions and using parity arithmetic. – Gone Mar 3 '12 at 23:59

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