Simplify complexity of n multichoose k
Edit: This answer is for $k&lt;n$. Without bounding $k$ in terms of $n$ the expression is unbounded.

If $k=n-1$ then your expression becomes 
$O\left ({{2(n-1)}\choose{n-1}}\right)$. Notice that by Stirling's formula for any $0&lt;\alpha&lt;1$
$${m \choose {\alpha m}}= \Theta(m^{-{1/2}} 2^{H(\alpha)m}),$$
where $H(q)=-q \log q - (1-q) \log (1-q)$ is the binary entropy. In particular $H(1/2)=1$. Therefore we have for $k=n-1$
$$O\left ({{2(n-1)}\choose{n-1}}\right)=\Theta((2n-2)^{-{1/2}} 2^{2n-2})=\Theta\left(\frac{ 4^{n}}{\sqrt{n}}\right).$$

Since the upper bound $k=n-1$ is the worst case (
I leave it as an exercise to show this), your expression is $O\left(\frac{ 4^{n}}{\sqrt{n}}\right)$.