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I have a proof that given a partition $\lambda=(\lambda_1,\dots,\lambda_l)$ then the number of semi-standard Young tableaux (SSYT) 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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    $\begingroup$ 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). ... $\endgroup$ Sep 7, 2012 at 14:00
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    $\begingroup$ ... 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. $\endgroup$ Sep 7, 2012 at 14:02
  • $\begingroup$ @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. $\endgroup$ Sep 7, 2012 at 14:03
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    $\begingroup$ 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. $\endgroup$ Sep 8, 2012 at 0:28
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    $\begingroup$ I believe that this formula is due to MacMahon. $\endgroup$
    – Ira Gessel
    May 23, 2014 at 22:18

2 Answers 2

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So, to answer this question, the formula is known. It is also (now) on the wikipedia page about Schur polynomials, so the next person will find it easily.

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I've finally found a place that gives this formula explicitly, without any need for piecing together or rewriting or taking limits using L'Hopital: Martin Aigner, A Course in Enumeration, Springer 2007, Section 8.5, formula (7) and the sentence that follows it.

I am wondering if anyone has ever given a bijective proof, though (after multiplying with the denominator).

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  • $\begingroup$ That's a nice reference - it had some other chapters which happened to be relevant for a small project I might be working on $\endgroup$ Apr 23, 2020 at 13:44

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