Let $k,m$ and $r$ be positive integers. 


Define 
$$\Omega(k,m,r) = \binom k {m-2r}\binom {k-m+2r} r$$
and
$$\Omega(k,m) = \sum\_{r=\max\{0,m-k\}}^{[\frac{m}{2}]}\Omega(k,m,r).$$



> 
> **Question.**
>  1. Is $\Omega(k,m)$ has a simple formula? 
>  2. Is $\frac{\sum\_{r=\max\{0,m-k\}}^{[\frac{m}{2}]}r\Omega(k,m,r)}{(k+m)\Omega(k,m)}$ has a simple formula? 
> 
> 
> 


**Background:**
If we want to calculate the numbers of $m$ white balls insert in $k-1$ black balls with the constraint that continuum three white balls $\circ\circ\circ$ is forbidden, then we can derive the formula above.



simple formula for a finite sum of multinomial numbers
First notice that
$$\Omega(k,m,r) = \binom{k}{m-2r,r,k-m+r}.$$
It follows that $\Omega(k,m)$ equals the coefficient of $x^{m-k}$ in $(1+x+x^{-1})^k$, which is the same as the coefficient of $x^{m}$ in $(1+x+x^2)^k$, also known as the [trinomial coefficient](https://en.wikipedia.org/wiki/Trinomial_triangle) $\binom{k}{m-k}\_2$. That is, $\Omega(k,m) = \binom{k}{m-k}\_2$.


The sum $\sum\_r r\Omega(k,m,r)$ can be obtained by differentiating $(1+xy+x^{-1})^k$ at $y=1$, which yields $k\binom{k-1}{m-k-1}\_2$.


