- A partition $\lambda$ of the number $n$.
- An assignment of numbers $\{1, 2, \dots n\}$ onto the diagram of the partitionsuch that the assignment is (a) weakly increasing in the rows, and(b) strictly increasing in the columns. It is strictly increasing in the columnsbecause gravity acts downwards.
- Formally, a partition is written as $\lambda \equiv [\lambda_1, \lambda_2, \dots, \lambda_m]$, where $\lambda_i \geq 0$ and $\sum_i \lambda_i = n$, and that theyare weakly decreasing ($\lambda_1 \geq \lambda_2 \geq \dots$).
- Formally, to define the tableaux, we first define the diagram $dg(\lambda) \equiv \{ (i, j) : 1 \leq j \leq \lambda[i] \}$which are the "locations" of the cells when visualizing $\lambda$ as a Ferrers diagram.
- Finally, the actual assignment of the numbers to the tableaux is given by a bijection$asgn: dg(\lambda) \rightarrow [n]$ such that $f$ is weakly increasing in the rowsand strictly increasing in the columns.

$n!/\left(\prod_{\texttt{cell} \in \lambda} hooklen(\texttt{cell})\right)$

where $hooklen$ is the largest "hook shape":
```
* * *
*
*
...
```

at the cell $(i, j)$ that is in the partition $\lambda$.
```
a b c d
e
f
```

And the numbers $\{1, 2, 3, 4, 5, 6\}$. How many ways can we assign the numbers
to the above hook shape such that its a legal young tableaux?
- First, see that $a = 1$ is a must. Proof by contradiction. Assume $1$is not placed at $a$. Whenever it is placed, it will be less than the numberplaced at $1$. But this is wrong, because a young tableaux must be weaklyincreasing in the rows and strictly increasing in the columns.
- Next, see that after placing $a = 1$, the other numbers can be placed "freely":If we take a subset of $\{2, 3, 4, 5, 6\}$ of the size of the leftover row,ie, $|b~c~d| = 3$, then there is only one way to place them such that they arein ascending order. Similarly, the leftover numbers go to the column where thereis only one way to place them.
- Hence, after $a = 1$ is fixed, for every $5C3$ subset, we get a legal hook.
- In general, if we have $n=r+c+1$ nodes, with $r+1$ nodes in the row, and $c+1$nodes in the column (the top-left node is counted twice), then we have $\binom{r+c}{r}$number of legal hooks; Once we pick the lowest number for the top left node,every $r$-subset will give us a hook.

- This result matches the hook-length formula. According to the hook lengthformula, we need to fill in for each node the length of the hook, and dividethe full product by $n!$. So forthe hook:

```
a b c d
e
f
```

This becomes:
```
6 3 2 1
2
1
```

$6!/(6\times 3!\times 2!) = 5!/(3! 2!) = 5C3 = \binom{r+c}{r}$

where $r=3, c=2$.
- Consider the hook shape. The only constraint we have is that the top-leftnumber ought to be the smallest. For the hook $H$ to be legal, if we distributenumbers into it uniformly at random, then there is a$1/(\texttt{hook-length}(H))$ probability that the hook will be legal.

- The tableaux will be legal iff
*all the hooks in the tableaux are legal*

- Thus, the probability of getting a legal tableaux is:

$\begin{aligned}
&\texttt{num}(\lambda)/n! = \prod_\{h \in \texttt{hook}(\lambda) 1/\texttt{hook-length}(h) \\
&\texttt{num}(\lambda) = n!/\prod_\{h \in \texttt{hook}(\lambda)\texttt{hook-length}(h) \\
\end{aligned}$

$RSK \equiv \bigcup_{\lambda \in \texttt{partition}(n)} SYT(\lambda) \times SYT(\lambda)$

given by the pair of insertion tableaux and the recording tableaux for each partition
$\lambda$ of $n$.
If we look at this in terms of set sizes, then it tells us that:
$\begin{aligned}
&|S_n| = |\bigcup_{\lambda \in \texttt{partition}(n)} SYT(\lambda) \times SYT(\lambda) \\
&n! = \sum_{\lambda \in \texttt{partition}(n)} |SYT(\lambda)|^2 \\
&n! = \sum_{\lambda \in \texttt{partition}(n)} |\texttt{hook-length-formula}(\lambda)|^2 \\
\end{aligned}$

This looks very suspicious, almost like the representation theoretic formula of:
$\texttt{group-size} = \sum_{\texttt{irrep} \in Repr(G)} dim(\texttt{irrep})^2$

and it is indeed true that $\texttt{hook-length-formula}(\lambda)$ corresponds
to the dimension of an irreducible representation of $S_n$, and each $\lambda$
corresponds to an irrep of $S_n$.