Probability of $l$ red columns when all $m$ rows have $k$ red and $n-k$ white balls Given: 


*

*There are $m$ rows of balls. 

*Each row has $k$ red balls and $n-k$ white balls, randomly placed. 


With what probability do there exist $l$ totally red collumns?

Update: The only thing I found out is the total number of possibilities:
$${n\choose k}^m$$
 A: Hint: I think the easiest way is to use the generalized principle of inclusion exclusion. Specifically, if there are several events $E_1,E_2,\dots,E_n$, and we let $X$ denote the number of these events which occur, then
$$
P(X\ge \ell) = \sum_{j=\ell }^n (-1)^{j-\ell}\binom{j-1}{\ell-1}\sum_{S\subseteq [n],|S|=j} P(E_S)\tag 1
$$
where $S$ ranges over all subsets of $[n]\stackrel{\text{def}}=\{1,2,\dots,n\}$ of size $j$, and $E_S=\bigcap_{i\in S}E_i$. 
Specifically, $E_i$ be the event that the $i^{th}$ column is all red. Then $P(E_S)$ is the event that all columns in $S$ are red, which is not hard to compute. Plugging that answer into $(1)$, you get a summation formula for the probability that at least $\ell$ red columns exist. 
Alternatively, the probability that exactly $\ell$ columns are red is 
$$
P(X=\ell) = \sum_{j=\ell }^n (-1)^{j-\ell}\binom{j}{\ell}\sum_{S\subseteq [n],|S|=j} P(E_S)\tag 2
$$
See this answer for a proof of $(2)$, from which you can derive $(1)$ by summing (after applying Pascal's identity, there will be a telescopic sum). 
