I've come to ask a question about the hook-length formula for the number of stadard Young tableaux given a Ferrer's diagram \mathcal{F}. But first I feel like I owe it to the interested outsider to provide some background:

Background

The beginnings of why most people would care about my eventual question lies in the depths of representation theory of finite groups (rep theory). Roughly the goal of rep theory is to study homomorphisms of the form \rho:G\to\text{GL}(V) where G is some finite group and \text{GL}(V) is the general linear group over some complex vector space V called the representation for \rho (one can study representations on vector spaces over fields other than \mathbb{C}, but things get much messier--especially in characteristic p).One of the most easily understood reason why anyone would want to study representation theory is that it often enables one to give slicker proofs of some theorems in group theory--often the alternative pure group theoretic proof being much, much more difficult. The classical example if this being Burnside's Theorem


One discovers that (like most math subjects) the study of all representations is redundant--one can study more fundamental building blocks to get the idea about everything (like studying simple groups instead of general groups). In representation theory these building blocks take the form of irreducible representations which are homomorphisms \rho:G\to\text{GL}(V) such that the only subspaces of V which are invariant under \rho_g=\rho(g) for every g\in G is \bold{0} and V. More fundamental though than the irreps themselves are the equivalence classes of irreps where an equivalence relation is defined on the set of all representations of G (known as \text{Rep}(G)) by declaring that \rho:G\to\text{GL}(V) and \psi:G\to\text{GL}(W) are equivalent if there exists an isomorphism T:V\to W such that T\rho_g=\psi_g T for every g\in G. There are some fascinating theorems about these equivalence classes. To name some it's true that the number of equivalence classes of irreps of G is the number of conjugacy classes of G, if \widehat{G} denotes the set of all these equivalence classes and for each \alpha\in\widehat{G} d_\alpha denotes the common dimension for each of the representation spaces for each \rho\in\alpha then \displaystyle \sum_{\alpha\in\widehat{G}}d_\alpha^2=|G|, the number of equivalence classes in \widehat{G} with d_\alpha=1 is equal to \left|G^{\text{ab}\right| (where G^{\text{ab}} is the abelinization of G defined to be G/[G,G]), d_\alpha\mid |G|, etc.


Of course one of the most interesting groups to study is the symmetric group S_n and the awesome fact is that there is a beautiful interplay between combinatorics and the study of the representation theory of S_n. To understand this one needs to understand what a Ferrer's diagram is. Namely, a Ferrer's diagram is a collection of left-justified boxes arranged in rows such that (going top to bottom) each row has no less boxes than the one succeeding it. For example:






(sorry for the gray background--it was taken from my blog). We normally denote Ferrer's diagrams with symbol like \mathcal{F} and if the diagram as n boxes we're apt to call it a n-frame. From a Ferrer's diagram we can define a Young tableaux to be a n-frame with the boxes bijectively filled in with the numbers \{1,\cdots,n\}. So, for example




Much more important for our purposes though are standard Young tableaux which are Young Tableaux in such a way that the columns and rows are increasing as one travels down and to the right respectively. So while the above Young tableaux is not standard the following one is




An interesting question then is that given a n-frame \mathcal{F} how many standard Young tableaux are there whose underlying Ferrer's diagram is \mathcal{F}? We denote the number of these by f_{\text{st}}\left(\mathcal{F}\right).


So, what does this have to do with representation theory? Well, it turns out that if tex \text{Frame}_n denotes the set of all n-frames that there exists a bijection \text{Frame}_n\to\widehat{S_n} (recalling that \widehat{S_n} is the equivalence classes of irreps of S_n) and moreover that the correspondence is such that if \rho^{(\mathcal{F})}:S_n\to\text{GL}(V) is the irrep associated to \mathcal{F}\in\text{Frame}_n then \dim_\mathbb{C} V=f_{\text{st}}\left(\mathcal{F}\right) and since these dimensions are important to us (as the previous list of facts seems to suggest) this is great!....assuming that there is an easy way to calculate f_{\text{st}}\left(\mathcal{F}\right). Fortunately there is-- a very easy method! To describe this formula (yes it is a formula) one needs to define the hook-length of a n-frame. Namely, given a n-frame \mathcal{F} and some (i,j)\in\mathcal{F} (where (i,j) denotes the block at the (i,j)^{\text{th}} position--using the usual matrix notation) we define the hook length of (i,j) denoted h_{i,j} to be the number of blocks to the right of (i,j) plus the number of blocks below (i,j) plus one. In practice this counts the the 'hook' formed by picking a position (i,j) and coloring all the blocks in that position's row to the right and the position itself and coloring the boxes below the position and counting the colored block. For example, the hook at position (3,2) in the following Ferrer's diagram is marked





It turns out then that for a given n-frame \mathcal{F} one has that



\displaystyle f_{\text{st}}\left(\mathcal{F}\right)=\frac{n!}{ \displaystyle \prod_{(i,j)\in\mathcal{F}}h_{i,j}}





The Question


Why is this true? I can prove it using an admittedly ineqloquent method, but I have zero intuition about it. Does anyone know a logical explanation of why this makes sense?