Given two integers $n,k>0$ such that $k\le n$ , we have the following inequalities for the binomial coefficient ${n\choose k}$ : \begin{eqnarray*} {n \choose k} & \le & \frac{n^k}{k!} \\ {n \choose k} & \le & \left(\frac{n\cdot e}{k}\right)^k \\ {n \choose k} & \ge & \left(\frac{n}{k}\right)^k \\ \end{eqnarray*}Here $e$ is the base of natural logarithms. Also, for large $n$ , ${n \choose k} \approx \frac{n^k}{k!}$ .