$k$th derivative with respect to $y$ of $(x+y)^j y^{n-j}$ Let $n \ge k \ge j \ge 1$ and $x,y$ are real variable. What is
$$
\frac{\partial^k}{\partial y^k} ((x+y)^j y^{n-j})_{x=1,y=-1}?
$$
It seems it would be $(-1)^{n-k} k! \left( n-j \atop k-j \right)$.
 A: Observe the following: in order for a term in the final answer to have a nonzero value, each $x+y$ term must be differentiated exactly once (and each $y$ term must be differentiated at most once, so that $k-j$ of the $y$ terms must be differentiated exactly once).
There are $\binom{n-j}{k-j}$ ways of picking which $y$ terms get differentiated, and for each of these, $k!$ orders possible for differentiating the $(x+y)$ and selected $y$ terms. Each of these has $n-j - (k-j)$ leftover undifferentiated $y$ terms, which each contribute $-1$ when evaluated at $y=-1$. Therefore the answer is
$$\binom{n-j}{k-j}k!(-1)^{n-k},$$
as you conjectured.
A: I guess @user7530 pointed out the right direction, but I was writing this lengthy answer, so I'll write it down here just in case it is helpful:
For simplicity, we use $a, b$ in place of $x+y, y$. Using product rule, we find that the $j$-th derivative with respect to $y$ is
$$j!a^0b^{n-j}+\binom{j}{1}(j\cdots 2)(n-j)a^1b^{n-j-1}+\binom{j}{2}(j\cdots 3)(n-j)(n-j-1)a^2b^{n-j-2}+\cdots+\binom{j}{j-1}j(n-j)\cdots(n-2j+2)a^{j-1}b^{n-2j+1}+(n-j)\cdots(n-2j+1)a^jb^{n-2j}.$$
Now we find $j+1, j+2,\dots$-th derivative. Notice that when evaluated at $x=1,y=-1$, the terms with $a$ disappear. So we try to find a pattern and I will not write down all the terms. 
The $j+1$-th derivative:
$$j!(n-j)b^{n-j-1}+\binom{j}{1}(j!)(n-j)b^{n-j-1}+\binom{j}{1}(j\cdots 2)(n-j)(n-j-1)ab^{n-j-2}+\binom{j}{2}(j\cdots 2)(n-j)(n-j-1)a b^{n-j-2}+\binom{j}{2}(j\cdots 3)(n-j)(n-j-1)(n-j-2)b^{n-j-3}+\cdots$$
Notice that the first two terms combine as $(j+1)!(n-j)b^{n-j-1}$, the next one combines as 
$$\binom{j+2}{2}(j\cdots 2)(n-j)(n-j-1)ab^{n-j-2}$$
The latter ones combine as $\binom{j+1}{3}\cdots$, where we use Pascal's identity 
$$\binom{n-1}{k-1}+\binom{n-1}{k}=\binom{n}{k}.$$
Now doing the same thing for $j+2$-th derivative, we obtain
$$(j+1)!(n-j)(n-j-1)b^{n-j-2}+\binom{j+1}{2}(j\cdots 2)(n-j)(n-j-1)b^{n-j-2}+\cdots.$$
The first two term combine as
$$\frac{1}{2}(j+2)!(n-j)(n-j-1)b^{n-j-2}.$$
Using Pascal's identity again we see the other terms has $\binom{j+2}{i}$ as coefficients. Repeating this process, we obtain
$$\frac{1}{(k-j)!} k! \frac{(n-j)!}{(n-k)!}(-1)^{n-k}$$
when plugging $0$ into $a$ and $-1$ into $b$.
A: 
One can expand $(x+y)^j$ and then take $k$-th derivative with respect
  to $y$. But soon the calculation becomes messy.

Well carrying out this, we have
$$
\frac{\partial^k}{\partial y^k}(1+y)^j y^{n-j} = \frac{\partial^k}{\partial y^k}\sum_{r=0}^j \binom{j}{r}y^{n-j+r} = \sum_{r=0}^j \binom{j}{r}\frac{\partial^k}{\partial y^k}y^{n-j+r}
$$
Now applying the derivative on $y^m$ and rearranging the terms a bit
\begin{align}
\sum_{r=0}^j \binom{j}{r}\frac{\partial^k}{\partial y^k}y^{n-j+r}&=\sum_{r=k+j-n}^j \binom{j}{r} \frac{(n-j+r)!}{(n-j+r-k)!} y^{n-j+r-k}\\
&=k!y^{n-k}\sum_{r=k+j-n}^j \binom{j}{r} \binom{n-j+r}{k} y^{-j+r}\\
&=k!y^{n-k}\sum_{r=0}^{j} \binom{j}{r} \binom{n-r}{k} y^{-r}
\end{align}
Since we are interested in $y=-1$, we reduced the original conjecture to 
$$
\sum_{r=0}^{j} \binom{j}{r} \binom{n-r}{k} (-1)^{r} = \binom{n-j}{k-j}.
$$
I could'nt find purely algebraic proof of the last proposition, but it has known combinatorial proof, see for example Use the Inclusion-Exclusion Principle to show that for any non-negative integers $m, r, n$ such that $m≤r≤n$.... 
