Suppose that $N$ and $r$ are positive integers. Prove or disprove that if $N$ is an even integer and $r$ is odd, then $\binom{N}{r}$ is even.
My attempt:
Let $N=2m$ and $r=2k+1$. Then $$\binom{N}{r}=\binom{2m}{2k+1}=\dfrac{(2m)!}{(2(m-k)-1)!(2k+1)!}$$
Also, we know that $\binom{N}{r}$ is always an integer. How do I proceed to show it is even or odd?
Also note that $n!$ is even for all $n\ge2$.
Best Answer
By Legendre's formula, the power of $2$ dividing $n!$ is $$v_2(n!)=\sum_{i=1}^\infty \left\lfloor\frac{n}{2^i}\right\rfloor.$$ So the power of $2$ dividing $\binom{n}{r}=\frac{n!}{r!(n-r)!}$ is $$\sum_{i=1}^\infty\left( \left\lfloor\frac{n}{2^i}\right\rfloor-\left\lfloor\frac{r}{2^i}\right\rfloor-\left\lfloor\frac{n-r}{2^i}\right\rfloor\right).$$
For each $i$, the term in parentheses is nonnegative since $\lfloor a\rfloor + \lfloor b\rfloor \leq \lfloor a+b\rfloor$ for any $a,b$. Do you see why the term for $i=1$ must be strictly positive, which would imply that the sum is also positive and give the conclusion you want?