Hook-length formula

Statement

Suppose $\lambda$ is an unordered integer partition of a natural number $n$. We can describe $\lambda$ by means of a Young diagram (also called a Ferrers diagram).

For every box in this Young diagram, define the hook-length at that box as follows:

Hook-length = 1 + (number of boxes directly below it vertically) + (number of boxes directly to the right of it horizontally)

Then, we have the following:

Degree of irreducible representation corresponding to partition $\lambda$ = Number of (standard) Young tableaux with shape $\lambda$ = Number of paths in the Young lattice from the trivial partition of $1$ to $\lambda$ = $\frac{n!}{\prod \operatorname{hook-lengths}}$

The product in the denominator is the product, over all boxes in the Young diagram, of the hook-length corresponding to that box.

Particular cases

$n$	$\lambda$	Young diagram	Hook lengths	Hook-length formula	Result of formula	Cross-check
1	1	PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.	1 (for unique box)	$\frac{1!}{1}$	1	unique Young tableau: fill box with $1$
2	2		2 (for left box), 1 (for right box)	$\frac{2!}{2 \cdot 1}$	1	unique Young tableau: 1 in left box, 2 in right box
2	1 + 1		2 (for top box), 1 (for bottom box)	$\frac{2!}{2 \cdot 1}$	1	unique Young tableau: 1 in top box, 2 in bottom box
3	3		3 (for left box), 2 (for middle box), 1 (for right box)	$\frac{3!}{3 \cdot 2 \cdot 1}$	1	unique Young tableau: 1 in left, 2 in middle, 3 in right box
3	1 + 1 + 1		3 (for top box), 2 (for middle box), 1 (for right box)	$\frac{3!}{3 \cdot 2 \cdot 1}$	1	unique Young tableau: 1 in top, 2 in middle, 3 in bottom box
3	2 + 1		3 (for top left box), 1 (for right box), 1 (for bottom box)	$\frac{3!}{3 \cdot 1 \cdot 1}$	2	Two possibilities for the Young tableau: 1 in top left, 2 in right, 3 in bottom 1 in top left, 3 in right, 2 in bottom
4	4		4,3,2,1 as we move from left-most to right-most box	$\frac{4!}{4 \cdot 3 \cdot 2 \cdot 1}$	1	unique Young tableau: 1,2,3,4 filled left to right