Binomial coefficients in Geometric summation Guys please help me find the sum given below.
$$\sum_{k=j}^i\binom{i}{k}\binom{k}{j}\cdot 2^{k-j}$$
(NOTE):The two coefficients are multiplied by 2 power (k-j)
I am using the formula: $\binom{r}{m}\binom{m}{q}=\binom{r}{q}\binom{r-q}{m-q}$
But not able to reach something fruitful.
Thanks in advance.
 A: If we set $N=i-j$,
$$\begin{eqnarray*}\sum_{k=j}^{i}\binom{i}{k}\binom{k}{j}2^{k-j}=\frac{i!}{j!}\sum_{k=j}^{i}\frac{2^{k-j}}{(i-k)!(k-j)!}&=&\frac{i!}{j!}\sum_{k=0}^{N}\frac{2^k}{(N-k)!k!}\\&=&\frac{i!}{j!N!}\sum_{k=0}^{N}\binom{N}{k}2^k\\&=&\color{red}{\binom{i}{j}3^{i-j}}.\end{eqnarray*}$$
A: HINT: Start with a box of $i$ balls numbered from $1$ through $i$, so that they can be individually identified. There are $\binom{i}k$ ways to choose $k$ of them to move to a second box. There are then $\binom{k}j$ ways to choose $j$ of the balls in the second box to move to a third box. That leaves $k-j$ balls in the second box; there are $2^{k-j}$ ways to pick a subset of them to move to a fourth box. Thus, for each possible choice of $k$ there are 
$$\binom{i}k\binom{k}j2^{k-j}$$
possible different outcomes. Clearly this is possible if and only if $j\le k\le i$. If you sum over all of these possible values of $k$, you count the ways to distribute $i$ white balls amongst four boxes in such a way that the third box has exactly $j$ balls in it.


*

*Can you find a closed expression in terms of $i$ and $j$ for the number of ways to distribute $i$ white balls amongst four boxes in such a way that the third box has exactly $j$ balls in it?

A: Suppose we are trying to evaluate
$$\sum_{k=j}^n {n\choose k} {k\choose j} 2^{k-j}.$$
which is
$$2^{-j} \sum_{k=j}^n {n\choose k} {k\choose j} 2^{k}.$$
Introduce
$${k\choose j} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{k}}{z^{j+1}} \; dz$$
which has the property that it is zero when $k\lt j,$ so we may extend
$k$ back to zero to get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{j+1}} 
\sum_{k=0}^n {n\choose k} 2^k (1+z)^k
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{j+1}} 
\left(1+2(1+z)\right)^n
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{j+1}} 
\left(2z+3\right)^n
\; dz.$$
Extracting the coefficient we get
$${n\choose j} 2^j 3^{n-j}$$
for a final answer of
$${n\choose j} 3^{n-j}.$$ 
I hope you'll permit me to include this MSE link where the above method is used in a more sophisticated manner.
A: If you use the formula you noted you are changing two variable binomials by just one variable and the other is constant:
$$\sum_{k=j}^i\binom{i}{k}\binom{k}{j}\cdot 2^{k-j}=\sum_{j\le k\le i}\binom{i}{j}\binom{i-j}{k-j}2^{k-j}=\binom{i}{j}\sum_{0\le k-j\le i-j}\binom{i-j}{k-j}2^{k-j}$$
So you can see now a sum from zero to the upper parameter of the binomial coefficient multiplied by an exponent with the variable $k-j$, what is the definition of binomial expansion. Then
$$\sum_{j\le k \le i}\binom{i}{k}\binom{k}{j}\cdot 2^{k-j}=\binom{i}{j}(1+2)^{i-j}$$
These manipulations are valid for everything because $k$ and $j$ are entire numbers by definition of the binomial coefficient, and this is the unique condition where $\binom{i}{k}\binom{k}{j}=\binom{i}{j}\binom{i-j}{k-j}$ holds.
A: Well you were totally in right direction
$\binom{i}{j}\cdot \sum_{k=j}^{k=i}$$\binom{i-j}{k-j}\cdot 2^{k-j}$ can be written as 
$\binom{i}{j} \cdot \sum_{e=0}^{e=i-j}$$\binom{i-j}{e} \cdot 2^e$ .Apply binomial theorem to get the answer $\binom ij \cdot (1+2)^{i-j}$
