Is \binom{n}{r} even for even n and odd r?

[mathgirl1]

Prove that if n is even and r is odd then $$\binom{n}{r}$$ is even.

Solution: I know I have these two equalities
$$\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}=\frac{n(n-1)...(n-r+1)}{r!}$$

Now if n is even and r is odd then (n-r+1) is even. So it seems that we will have at least one more even number on the numerator than is in the denominator thus making the whole thing even. However, I am not sure how to show this. Also, part a of this questions was to show $$\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$$ and my professor usually gives part b that uses part a. But I am not sure how to use part a to show this without using the definition. I know the only case we would have to look at is if one is even and one is odd and then show a contradiction I guess? But I didn't get very far with this approach. Any help is much appreciated.

Thanks in advance!

 

[johng1]

Hi mathgirl,
Here's a solution, but it does not use the additive property that you cited, except maybe this was used to show every binomial coefficient is an integer.

Let $n$ and $r$ be positive integers with $1\leq r\leq n$ (so $n-r+1\neq0$). Then
$$\binom{n}{r}={n!\over r!(n-r)!}={(n-r+1)n!\over r(r-1)!(n-r+1)!}={n-r+1\over r}\binom{n}{r-1}$$

Now assume $n$ is even and $r$ is odd. Then $n-r+1$ is even and from
$$r\binom{n}{r}=(n-r+1)\binom{n}{r-1}$$
it follows that
$$r\binom{n}{r}$$ is even. Since $r$ is odd, $$\binom{n}{r}$$ is even.