TITLE: Array of numbers, how many solutions/ways?
QUESTION [5 upvotes]: Let's say, we have an array/matrix $n \times m$ and we need to find, how many ways can we fill this array with numbers from $\{ 1, \ldots , m\cdot n \}$, but:
1) every number can be used only 1 time
2) every column and every row should be sorted in increasing order
For example, if we have $n = m = 3$, there are 42 different ways to do this.
I was thinking about this problem, but I can't find out any simple formula to compute this for every $n, m$. 
We can solve this problem by writnig a program that uses backtracking method/algorithm [1] but it has huge complexity and problem is solved by computer, and I want to know, how to do this by hand.
Sombody told me, that in this problem I can use some kind of Catalan numbers [2] [3], but I'm not sure about this method.
So, what do You think about this problem and its solution?
[1] Backtracking (Wikipedia): http://en.wikipedia.org/wiki/Backtracking 
[2] Catalan numbers (Wikipedia): http://en.wikipedia.org/wiki/Catalan_number 
[3] Catalan numbers (OEIS): http://oeis.org/A000108

REPLY [6 votes]: These are Young tableaux.  OEIS has the numbers.  For the square case, see A039622.   For the rectangular case, see A060854

REPLY [5 votes]: You are counting the number of standard Young tableaux of shape $(m, m, m, ...)$ (where $m$ occurs $n$ times).  The formula that counts these is called the hook length formula.  It correctly reproduces the answer
$$\frac{9!}{5 \cdot 4 \cdot 4 \cdot 3 \cdot 3 \cdot 3 \cdot 2 \cdot 2} = 42$$
for the example you gave.  The general formula seems annoying to write down.  In the special case $n = 2$ you get Catalan numbers; this is a nice exercise and it is worth trying to prove this with and then without the hook length formula.