Prove this identity with a combinatorial argument Prove that for all integers $j,k,n, k \ge 0$, we have
$$ \sum\limits_{i=0}^n ({-1})^{n-i} \binom {n}{i} \binom {ki}{j} = \begin{cases}
0, & \text{if } 0 \le j < n; \\
k^n, & \text{if } j = n.
\end{cases} $$
The alternating sum suggests use of inclusion-exclusion principle, but I am unable to setup  a situation. The sum becomes 0 which is strange for me. How do you incorporate that. Using inclusion -exclusion for counting something like surjective functions always we count something.
I am trying to learn combinatorics, so would prefer a combinatorial argument.  
 A: Consider $n$ groups of $k$ objects. The quantity ${n\choose i}{ki\choose j}$ represents the number of ways to choose $i$ of the groups, then select $j$ objects from the chosen groups. By inclusion-exclusion, the left hand side is the number of ways to choose $j$ objects total from a collection of $n$ groups of $k$ objects, such that each group has at least one object selected from it. It is not hard to see the right hand side counts the same thing.
A: This  problem  also  has  an   algebraic  proof  which  I  submit  for
enrichment. Suppose we seek to evaluate
$$\sum_{q=0}^n (-1)^{n-q} {n\choose q}
{kq\choose j}.$$
We introduce
$${kq\choose j} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{j+1}} (1+z)^{kq}
\; dz.$$
With $j$ non-negative this  integral correctly represents the binomial
coefficient and vanishes when $j$ is out of range. We then get for the
sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{j+1}}
\sum_{q=0}^n {n\choose q} (-1)^{n-q} (1+z)^{kq}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{j+1}}
(-1+(1+z)^k)^n
\; dz.$$
This is
$$[z^j] \left({k\choose 1}z + {k\choose 2}z^2 + \cdots +
{k\choose k}z^k\right)^n.$$
The first power of $z$ that occurs  here is $z^n$ so the value is zero
for $j\lt n.$  Moreover the coefficient on $z^n$ can  only be achieved
in  one  way,  namely by  taking  the  first  term  of the  sum  being
exponentiated. This  has coefficient  $k$ so the  answer for  $j=n$ is
$$k^n.$$
Remark. We can use this technique for larger values of $j.$ For
example we get for $j=n+2$ the result
$${n\choose 1} k^{n-1} {k\choose 3}
+ {n\choose 2} k^{n-2} {k\choose 2}^2.$$
