A Bernstein-like Combinatorial Sum Do sums of the "convoluted Bernstein" form 
$$
\sum_{j=1}^m {m\choose j} q^j (1-q)^{m-j}  {n \choose k} \left(\frac j m\right)^k \left(1- \frac j m\right)^{n-k} 
$$
admit closed forms? My attempts to use generating functions have not worked. 
 A: First off, it is always worth to remove common factors (not depending on the index of summation) from the summands. The given sum is reduced to (I also assume $k>0$ to have summation start from $j=0$):
$$\binom{n}{k} \frac{m!}{m^n}\cdot \sum_{j=0}^m \frac{q^j j^k}{j!}\cdot \frac{(1-q)^{m-j} (m-j)^{n-k}}{(m-j)!}.$$
Notice that $\frac{q^j j^k}{j!}$ is the coefficient of $x^j$ in $\left(x\frac{d}{dx}\right)^k e^{qx}$. 
Similarly, $\frac{(1-q)^{m-j} (m-j)^{n-k}}{(m-j)!}$ is the coefficient of $x^{m-j}$ in $\left(x\frac{d}{dx}\right)^{n-k} e^{(1-q)x}$.
Furthermore, it is helpful to know that 
$$\left(x\frac{d}{dx}\right)^t = \sum_{s=0}^t S(t,s)\cdot x^s\cdot \left(\frac{d}{dx}\right)^s,$$
where $S(t,s)$ are Stirling numbers of the second kind.
Combining, we have that $\frac{q^j j^k}{j!}$ is the coefficient of $x^j$ in $$\sum_{s=0}^k S(k,s)\cdot x^s\cdot q^s\cdot e^{qx} = B_k(qx)\cdot e^{qx},$$
where $B_k(x)$ is Bell polynomial of first kind (or Touchard polynomial).
Similarly, $\frac{(1-q)^{m-j} (m-j)^{n-k}}{(m-j)!}$ is the coefficient of $x^{m-j}$ in $B_{n-k}((1-q)x)\cdot e^{(1-q)x}$.
In summary, we have that the original sum (but starting at $j=0$) equals the coefficient of $x^m$ in
$$\binom{n}{k}\cdot \frac{m!}{m^n}\cdot B_k(qx)\cdot B_{n-k}((1-q)x)\cdot e^x.$$
UPD. It may be useful to know that
$$\sum_{k=0}^n \binom{n}{k}\cdot B_k(qx)\cdot B_{n-k}((1-q)x) = B_n(x).$$
