Consider the partition $\lambda=(m,n-m)$ of $n$ (thus $2m \ge n$).
The number of standard Young tableaux of shape $\lambda$ is given by $$f_{(m,n-m)} = \binom nm - \binom{n}{m+1}$$
a) Prove this using the hook-length formula.
b) Prove this using induction on $n$.
Please, help!
On
HINTS:
(a) This is straightforward algebra once you work out the hook-lengths. Do that in three parts: the cells in the second row (part $0$ in the diagram below); those in the first row with nothing below them (part $1$ in the diagram below); and those in the first row with a cell below them (part $2$ in the diagram below).
$$\begin{array}{|c|c|} \hline 2&2&2&\dots&2&1&1&\dots&1\\ \hline 0&0&0&\dots&0\\ \hline \end{array}$$
(b) This uses nothing more complicated than Pascal’s identity, once you see how the induction step works. In a Young tableau of shape $(m,n+1-m)$ there are only two places for the $n+1$: at the end of the first row, and at the end of the second row. If it’s at the end of the first row, it’s been added to a Young tableau of shape $(m-1,n-m+1)$; if it’s at the end of the second row, it’s been added to a Young tableau of shape $(m,n-m)$.
As I don't like doing as I'm told, here is a method that uses neither the hook length formula nor induction.
If you just choose $m\geq\frac n2$ numbers from $\{1,\ldots,n\}$ for the first row, arrange them there in increasing order, and put the remaining $n-m$ values in increasing order in the second row, then you obtain what I'll call a tabloid of shape $(m,n-m)$: rows are increasing but columns need not be. All tableaux of shape $(m,n-m)$ are among these tabloids, but there are non-tableaux as well. For an entry$~x$ of a tabloid, call its "overhang" the difference, in the sub-tabloid with entries${}\leq x$, of the length of its first row minus the length of its second row; a tabloid is a tableau precisely if it has no entries with overhang${}<0$. For any tabloid that is not a tableau, find the most negative occurring overhang, and the smallest entry $x$ with that overhang; taking $x$ out of its row, necessarily the second row, and inserting it at the proper place in the first row produces a tabloid of shape $(m+1,n-(m+1))$. Note that this operation does not change the overhang for any entries${}<x$, but it increases the overhang of all entries${}\geq x$ by$~2$. This means that after the change the value $x-1$ is the largest one with the smallest occurring (non-positive) overhang; assume the existence of a virtual entry $0$ with overhang $0$ to ensure this is well defined even in the case $x=1$. Then the operation has an inverse, that can be applied to any tabloid of shape $(m+1,n-(m+1))$; it consists of locating this value $x-1$ and thereby $x$, and moving $x$ from the first row to the second row.
All this shows that the number of tableaux of shape $(m,n-m)$ equals the number of tabloids of shape $(m,n-m)$ minus the number of tabloids of shape $(m+1,n-(m+1))$, which is $$ \binom nm-\binom n{m+1}. $$
Added. While I feel the above is in a sense the "best" way to bring non-tableau tabloids of shape $(a,b)$ in correspondence with tabloids of shape $(a+1,b-1)$, especially for understanding what happens when iterating the operation until a tableau of some shape $(a+d,b-d)$ is obtained, I realise that there is also a bijection that is somewhat easier to describe, based on a "reflection principle". To this end, find the smallest entry $x$ in the tabloid that violates the tableau condition by being in the second row below an entry $y>x$; now take the part of the first row strictly to the left of $y$ and exchange it with the part of the second row up to and including $x$ (shifting the remainder of the rows to accommodate the fact that the parts exchanged differ by$~1$ in length). The reverse operation locates $x$ as the smallest entry in the first row that is either below an entry${}>x$ (so not violating the tableau condition) or below an empty square; this must happen for some entry since the first row is strictly longer than the second one when the shape is $(a+1,b-1)$.