Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
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
add comment

1 Answer

up vote 5 down vote accepted

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.

share|improve this answer
    
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. –  Bill Dubuque Mar 3 '12 at 23:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.