フック長の公式
The ideas of Vershik–Kerov and Logan–Shepp were later refined by Jinho Baik, Percy Deift and Kurt Johansson, who were able to achieve a much more precise analysis of the limiting behavior of the maximal increasing subsequence length, proving an important result now known as the Baik–Deift–Johansson theorem. Their analysis again makes crucial use of the fact that d_\lambda has a number of good formulas, although instead of the hook length formula it made use of one of the determinantal expressions.
The formula for the number of Young tableau of a given shape was originally derived from the Frobenius determinant formula in connection to representation theory.  If the shape of a Young diagram is given by the row lengths n_1,\dots,n_m, then the number of tableau with that shape is given by
f(n_1,n_2,\ldots,n_m) = \frac{n! \,\Delta(n_m,n_{m-1}+1,\ldots,n_1 + m - 1)}{n_m ! (n_{m-1} + 1)! \cdots (n_1 + m - 1)!}
Hook lengths can also be used to give a product representation to the generating function for the number of reverse plane partitions of a given shape.  If λ is a partition of some integer p, a reverse plane partition of n with shape λ is obtained by filling in the boxes in the Young diagram with non-negative integers such that the entries add to n and are non-decreasing along each row and down each column.  The hook lengths  h_1,\dots,h_p  can be defined as with Young tableau.  If πn denotes the number of reverse plane partitions of n with shape λ, then the generating function can be written as
\sum_{n=0}^\infty \pi_n x^n = \prod_{k=1}^p (1-x^{h_k})^{-1}
Stanley discovered another formula for the same generating function. In general, if  A  is any poset with  n  elements, the generating function for reverse  A -partitions is
\frac{P(x)}{(1-x)(1-x^2)\cdots(1-x^n)}
where  P(x)   is a polynomial such that P(1) is the number of natural labelings of  A .
In the case of a partition  \lambda , we are considering the poset in its cells given by the relation
(i,j) \leq (i',j') \iff i\leq i' \qquad \textrm{and} \qquad j\leq j'.
So a natural labeling is simply a standard Young tableau, i.e.  P(1) = f^{\lambda}
Combining the two formulas for the generating functions we have
\frac{P(x)}{(1-x)(1-x^2)\cdots(1-x^n)}  = \prod_{(i,j)\in \lambda} (1-x^{h_{(i,j)})^{-1}
Both sides converge inside the disk of radius one and the following expression makes sense for  |x| < 1
P(x) = \frac{\prod_{k=1}^n (1-x^k)}{\prod_{(i,j)\in \lambda} (1-x^{h_{(i,j)})}.
It would be violent to plug in 1, but the right hand side is a continuous function inside the unit disk and a polynomial is continuous everywhere so at least we can say
P(1) = \lim_{x\to 1} \frac{\prod_{k=1}^n (1-x^k)}{\prod_{(i,j)\in \lambda} (1-x^{h_{(i,j)})}.
Finally, applying L'Hopital's rule  n  times yields the hook length formula
P(1) = \frac{n!}{\prod_{(i,j)\in \lambda}h_{(i,j)}.
Specializing the schur functions to the variables  1,t,t^2,t^3, \cdots  there is the formula
s_{\lambda}(1,t,t^2,\cdots) = \frac{t^{n\left( \lambda\right)}{\prod_{(i,j)\in Y(\lambda)}(1-t^{h_{\lambda}(i,j)})}
The number  n(\lambda)  is defined as
n(\lambda) = \sum_i (i-1)\lambda_i = \sum_i \binom{\lambda_i'}{2}
where  \lambda'  is the conjugate partition
There is a generalization of this formula for skew shapes,
s_{\lambda/\mu}(1,t,t^2,\cdots) = \sum_{S \in E(\lambda/\mu)} \prod_{(i,j) \in \lambda\setminus S} \frac{t^{\lambda_j'-i}{1-t^{h(i,j)}
where the sum is taken over excited diagrams of shape \lambda and boxes distributed according to \mu.
The Catalan numbers are ubiquitous in enumerative combinatorics. Not surprisingly, they are also part of this story:
C_n = f^{(n,n)}
Lets briefly mention why.  When doing a Dyck path we may go up (U) or down (D). So for any Dyck path of length  n  consider the tableaux of shape  (n,n)  such that the first row is given by the numbers  i  such that the  i -th step was up and in the second row given by the positions in which it goes down. For example, UUDDUD correspond to the tableaux with rows 125 and 346.
The hook formula gives another way of getting a closed formula for the Catalan numbers
C_n = \frac{(2n)!}{(n+1)(n)\cdots(3)(2)(n)(n-1)\cdots(2)(1)} = \frac{(2n)!}{(n+1)!n!} = \frac{1}{n+1}\binom{2n}{n}