Sum over j of Function of Floor of mj over n/Corollary

Corollary to Sum over $j$ of $\map f {\floor {\frac {m j} n} }$

 * $\ds \sum_{0 \mathop \le j \mathop < n} \binom {\floor {m j / n} + 1} k + \sum_{0 \mathop \le j \mathop < m} \ceiling {\dfrac {j n} m} \binom j {k - 1} = n \binom m k$

Proof
Hence: