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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.

2 Answers 2

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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.

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    ...except that I've now edited the Wikipedia page so that it says *standard* tableaux, and has a link to the hook-length formula.2011-03-09
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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.)