Hook length formula

Remarks

The formula is due to Frame, Robinson and Thrall. Let $F(\lambda)$ denote the right-hand side of the formula. A short proof by Greene, Nijenhuis and Wilf uses a random "hook walk" inside the diagram. They show that the walk ends at each corner $c$ with probability $F(\lambda \setminus c) / F(\lambda)$, so these ratios add up to $1$. The largest entry $n$ of a standard tableau sits in a corner. So $f^\lambda$ is the sum of $f^{\lambda \setminus c}$ over the corners $c$, and $F$ satisfies the same recursion; induction on $n$ finishes the proof.

Example: for $\lambda = (2, 2)$ the hook lengths are $3, 2$ in the first row and $2, 1$ in the second row. The formula gives $4! / (3 \cdot 2 \cdot 2 \cdot 1) = 2$, and indeed the standard tableaux are $\begin{smallmatrix} 1 & 2 \\ 3 & 4 \end{smallmatrix}$ and $\begin{smallmatrix} 1 & 3 \\ 2 & 4 \end{smallmatrix}$. For $\lambda = (n)$ the hook lengths are $n, n-1, \dots, 1$ and $f^\lambda = 1$.

Going deeper

For a student

The number $f^\lambda$ is the dimension of the irreducible representation of the symmetric group $S_n$ indexed by $\lambda$. So the hook length formula computes these dimensions directly. Together with the Robinson–Schensted correspondence, which gives $\sum_{\lambda} (f^\lambda)^2 = n!$, it is a basic tool of the representation theory of $S_n$. There are versions for shifted shapes and for trees, and a $q$-analogue that counts tableaux by a statistic.