Parity of Binomial Coefficients Do $\binom nr$ and $\binom {2n}{2r}$ always have the same parity? I can see that it's true for $r=1$ since $\binom {n}{1}=n$ and $\binom{2n}{2}=n(2n-1)$, but what about for bigget $r$?
 A: As per robjohn's suggestion:
$$\begin{align}
\binom{2n}{2r} &= \frac{2n}{2r}\times \left(\frac{2n-1}{2r-1}\right)\times \frac{2n-2}{2r-2}\times \left(\frac{2n-3}{2r-3}\right)\times \cdots \times \frac{2n-2r+2}{2}
\times \left(\frac{2n-2r+1}{1}\right)\\
&= \frac{n}{r}\times \frac{n-1}{r-1}\times\cdots\times \frac{n-r+1}{1} \times 
\left(\frac{2n-1}{2r-1}\right)\left(\frac{2n-3}{2r-3}\right)\cdots 
\left(\frac{2n-2r+1}{1}\right)\\
&= \binom{n}{r}\times \frac{2p+1}{2q+1}
\end{align}$$
where $p$ and $q$ are integers. Thus we have that
$$(2q+1)\binom{2n}{2r} = (2p+1)\binom{n}{r} \tag{1}$$
where the left side of $(1)$ is even or odd according as $\binom{2n}{2r}$ is
even or odd while the right side of $(1)$ is even or odd according as
$\binom{n}{r}$ is even or odd. Hence, $\binom{2n}{2r}$
and $\binom{n}{r}$ must have the same parity modulo 2.
More strongly, we can also deduce from $(1)$ that
the largest power of 2 that divides $\binom{2n}{2r}$
is the same as the largest power of 2 that divides $\binom{n}{r}$.
A: As shown in this answer, for any prime $p$, the number of factors of $p$ that divide $\binom{n}{k}$ is
$$
\frac{\sigma_p(k)+\sigma_p(n-k)-\sigma_p(n)}{p-1}
$$
where $\sigma_p(n)$ is the sum of the base-$p$ digits of $n$. Note that $\sigma_2(n)=\sigma_2(2n)$.
