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 $r$?

Solution 1:

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 $\binom{n}{r}$.

Solution 2:

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 $\sigma_2(n)=\sigma_2(2n)$.