<s>[INST] prove this by binomial theorem Prove that by binomial theorem :
$\dbinom{6}{0}+\dbinom{6}{2}+\dbinom{6}{4}+\dbinom{6}{6}=\dbinom{6}{1}+\dbinom{6}{3}+\dbinom{6}{5}$
I started to prove it by moving the right side to the left :
$\dbinom{6}{0}+\dbinom{6}{2}+\dbinom{6}{4}+\dbinom{6}{6}-\dbinom{6}{1}-\dbinom{6}{3}-\dbinom{6}{5}=0$ 
it should be done by the binomial coefficient ...?....
how ?? 
thanks in advance
 [/INST] HINT:
$$0=(1-1)^n=\sum_{r=0}^n\binom nr1^{n-r}(-1)^r=\sum_{r=0}^{2r\le n}\binom n{2r}-\sum_{r=0}^{2r+1\le n}\binom n{2r+1}$$
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