# New formula for counting SSYTs?

I have a proof that given a partition $\lambda=(\lambda_1,\dots,\lambda_l)$ then the number of semi-standard Young tableaux of shape $\lambda$ with entries in $1,2,\dots, n$ is given by

$$\frac{1}{1!2!\cdots (n-1)!} \prod_{1\leq i\lt j\leq n} (\lambda_i-i)-(\lambda_j-j).$$ (We define $\lambda_j:=0$ if $j \gt l.$)

The product is also recognized as a Vandermonde determinant.

There are plenty of product formulas (over boxes in the tableau) and determinant formulas (but not in Vandermonde form, as far as I can tell) for the number of such SSYTs, but I have not seen a this particular one in the literature or in any article I've come across.

Is this formula known? Is this formula of any interest?

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Probably known, sorry to disappoint. The number of semistandard Young tableaux of shape $\lambda$ is the value of the $\lambda$-Schur polynomial (which I consider as a polynomial in $n$ variables) at $\left(1,1,...,1\right)$ (because the $\lambda$-Schur polynomial is a generating function for semistandard Young tableaux of shape $\lambda$). But the value of the $\lambda$-Schur polynomial at $\left(1,1,...,1\right)$ is the dimension of the $\lambda$-Schur module of $\mathrm{GL}_n\mathbb C$ (since the $\lambda$-Schur polynomial is the character of this $\lambda$-Schur module). ... –  darij grinberg Sep 7 '12 at 14:00
... By Weyl's character formula (Theorem 4.63 in Pavel Etingof's representation theory notes www-math.mit.edu/~etingof/replect.pdf ), the latter dimension is $\prod\limits_{1\leq i < j \leq n} \dfrac{\lambda_i - \lambda_j + j - i}{j-i} = \dfrac{1}{1! 2! ... \left(n-1\right)!} \prod\limits_{1\leq i < j \leq n} \left(\left(\lambda_i - i\right) - \left(\lambda_j - j\right)\right)$. Combining these observations, we get your formula. I suspect that some text has this more explicitly. –  darij grinberg Sep 7 '12 at 14:02
@darij grinberg: Yes, I am aware that it is the evaluation of the Schur polynomial at (1,...1). Do you have a reference? I'd love to put this identity at Wikipedia for example, since, in my humble opinion, this is a very straightforward way to compute these numbers. –  Per Alexandersson Sep 7 '12 at 14:03
One reference is the case $q=1$ of Enumerative Combinatorics, vol. 2, (7.105) on page 375. This goes back to Littlewood and Richardson. See the discussion on page 403 of EC2. –  Richard Stanley Sep 8 '12 at 0:28
I believe that this formula is due to MacMahon. –  Ira Gessel May 23 '14 at 22:18