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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 $r$?

Nishant
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    Hint: $\displaystyle \binom{2n}{2r} = \frac{2n}{2r}\times \left(\frac{2n-1}{2r-1}\right)\times \frac{2n-2}{2r-2}\times \left(\frac{2n-3}{2r-3}\right)\times \cdots$ which should allow you to pull out a $\binom{n}{r}$ from that product on the right. Is everything else an odd number? – Dilip Sarwate Jan 24 '14 at 04:21
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    @DilipSarwate: you should make this into an answer. – robjohn Jan 24 '14 at 04:34

2 Answers2

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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 $\binom{n}{r}$.

Dilip Sarwate
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  • So you are saying $$\binom{2n}{2r}=\binom{n}{r}\frac{(2n-1)!!}{(2n-2r-1)!!(2r-1)!!}$$ and where $(2n-1)!!$, etc. are odd, and don't affect the parity of $\binom{2n}{2r}$. Good observation. – robjohn Jan 24 '14 at 08:57
  • @robjohn Thanks. I was going to edit my answer to make the final point easier to get at, but your comment beat me to the punch. I will give a slightly different approach anyway. – Dilip Sarwate Jan 24 '14 at 14:53
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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 $\sigma_2(n)=\sigma_2(2n)$.

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robjohn
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