From the wiki page Catalan number, we know the number of Young tableaux whose diagram is a 2-by-n rectangle given $2n$ distinct numbers is $C_n$. In general, given $m\times n$ distinct numbers, how many Young tableaux whose diagram is a $m\times n$ rectangle are there?
Also, what if these numbers can be repeated?
Many thanks.
For the answer to your main question, you need to use the hook-length formula.
OEIS A060854 gives the result $$(mn)! \prod_{i=0}^{n-1} \frac{i!}{(m+i)!} \textrm{ or equivalently } (mn)! \prod_{j=0}^{m-1} \frac{j!}{(n+j)!} $$ and some more information.
It not quite clear what you mean by allowing repeated numbers, but what one usually considers in that case is so-called semi-standard Young tableaux, i.e., tableaux which are increasing (strict inequality) down each column, but only nondecreasing (equality allowed) along each row. The number of such arrangements on a given Young diagram, where the numbers $1,2,\dots,N$ are allowed, is counted as follows: define the "content" of box $(i,j)$ in the diagram to be $i-j$. Here's an illustration:
0 1 2 3 4
-1 0 1 2
-2 -1 0
-3 -2
-4 -3
-5
Hook lengths are defined as for the usual hook-length formula for counting standard Young tableaux:
10 8 5 3 1
8 6 3 1
6 4 1
4 2
3 1
1
To get the answer, take the product over all boxes of (($N$ plus the content of that box) divided by (the hook length for that box)).
(This is a special case of something called Stanley's hook-content formula.)
