TITLE: A bijection between sets of Young tableaux of two kinds
QUESTION [27 upvotes]: Assuming that the problem of exhibiting a bijection is not considerd a frivolous pursuit, allow me to ask a question troubling me for some time now.
Let $\lambda \vdash n$ denote the fact that $\lambda$ is a partition of $n$. Denote the number of parts by $l(\lambda)$. If $T$ is a standard Young tableau (SYT), we will denote the underlying partition shape by $sh(T)$.
Given a positive even integer $2n$, let 
$$ Pe_{2n}=\{ \lambda: \lambda\vdash 2n,\text{ } l(\lambda) \leq3 \text{ and all parts of } \lambda \text{ are even} \}$$
and
$$ Qe_{2n}=\{ \lambda: \lambda\vdash 2n, \lambda = (k,k,1^{2n-2k}), \text{ }k\geq 1 \}$$
Using these sets we will define two more sets whose elements are SYTs.
$$ TPe_{2n}=\{T: T \text{ an SYT, } sh(T)\in Pe_{2n} \}$$
and
$$ TQe_{2n}=\{T: T \text{ an SYT, } sh(T)\in Qe_{2n} \}$$
$\textbf{Question}$:
Is there a bijective proof exhibiting the fact that the cardinalities of $TPe_{2n}$ and $TQe_{2n}$ are equal?
The second question is very similar. Given an odd positive integer $2n+1$, let
$$ Po_{2n+1}=\{ \lambda: \lambda\vdash 2n+1,\text{ } l(\lambda)=3 \text{ and all parts of } \lambda \text{ are odd} \}$$
and
$$ Qo_{2n+1}=\{ \lambda: \lambda\vdash 2n+1, \lambda = (k,k,1^{2n+1-2k}), \text{ }k\geq 1 \}$$
Using these sets we will define two more sets whose elements are SYTs.
$$ TPo_{2n+1}=\{T: T \text{ an SYT, } sh(T)\in Po_{2n+1} \}$$
and
$$ TQo_{2n+1}=\{T: T \text{ an SYT, } sh(T)\in Qo_{2n+1} \}$$
$\textbf{Question}$:
Is there a bijective proof exhibiting the fact that the cardinalities of $TPo_{2n+1}$ and $TQo_{2n+1}$ are equal?
I tried quite a few approaches ( Motzkin path interpretations, matching diagrams etc) but did not succeed. I hope somebody here can guide me.
The relevant OEIS entry would be link text
Thanks,
Vasu

REPLY [10 votes]: Let me explain a way to prove such results. There are (mostly) two types of people out there, those that think of the $n$th Catalan number as $\frac{1}{n+1}\binom{2n}{n}$ and those that think of it as $\binom{2n}{n}-\binom{2n}{n-1}$. The first type of people will likely think of Dyck paths as eqivalence classes or $\mathbb Z/(n+1)\mathbb Z$ orbits of paths, whereas the second type will probably think of Dyck paths as "paths inside an $n\times n$ rectangle that don't cross the origin" (See proof 2 vs proof 3 here). While the first type would probably be happy to count standard young tableaux using the usual form of the hook-length formula, the second type would probably prefer the (equivalent):
Jacobi-Trudi Formula
The number of Standard Young Tableaux on a partition of shape $\lambda=(\lambda_1,\dots,\lambda_k)$ of $n$ is
$$f^{\lambda}=\sum_{\sigma \in S_k}\varepsilon(\sigma)\binom{n}{\lambda_1+k-\sigma(k)\, ,\dots,\,\lambda_k+1-\sigma(1)}.$$
Where $\varepsilon$ denotes the sign of a permutation. Similar to the Catalan numbers in this presentation one sacrifices obvious positivity but has obvious integrality. For the sake of illustration, in the case $k=3$, this says
$$f^{(a,b,c)}=\binom{n}{a,b,c}+\binom{n}{a+1,b+1,c-2}+\binom{n}{a+2,b-1,c-1}-\binom{n}{a+1,b-1,c}-\binom{n}{a,b+1,c-1}-\binom{n}{a+2,b,c-2}.$$
The computation:
Let's take the case of three even parts, although this method will work for the odd case as well as the general case of partitions with a fixed (arbitrary) number of rows. If one sums the quantity $f^{(a,b,c)}$ over all triples of integers $a\geq b\geq c\geq 0$ with $a+b+c=n$ then the multinomial coefficients form the formula above telescope and we are left with
$$|TPe_{2n}|=\sum_{k\geq 0}\binom{2n}{k,k,2n-2k}-\sum_{k\geq 0}\binom{2n}{k,k+1,2n-2k-1}$$
Now we can turn our attention to $TQe_{2n}$. We can count the number of SYT of shape $(k,k,1^{2n-2k})$ from the hook length formula, but instead of presenting it in "type one" as Brian Hopkins did in the other answer, we will use the "type two" presentation $$f^{(k,k,1^{2n-2k})}=\binom{2n}{k}\binom{2n-k-1}{k-1}-\binom{2n}{k+1}\binom{2n-k}{k}$$
$$=\binom{2n}{k,k,2n-2k}-\binom{2n}{k}\binom{2n-k-1}{k}-\binom{2n}{k-1}\binom{2n-k}{k}.$$
So in order to get the desired equality we sum over all $k\geq 1$ 
$$|TQe_{2n}|=\sum_{k\geq 1}f^{(k,k,1^{2n-2k})}=\sum_{k\geq 1}\left(\binom{2n}{k,k,2n-2k}-\binom{2n}{k}\binom{2n-k-1}{k}-\binom{2n}{k-1}\binom{2n-k}{k}\right)$$
$$=\sum_{k\geq 1}\binom{2n}{k,k,2n-2k}-\sum_{k\geq 0}\left(\binom{2n}{k}\binom{2n-k-1}{k}+\binom{2n}{k}\binom{2n-k-1}{k+1}\right)+\binom{2n}{0}\binom{2n-1}{0}$$
$$=\sum_{k\geq 1}\binom{2n}{k,k,2n-2k}-\binom{2n}{k,k+1,2n-2k-1}+\binom{2n}{0,0,2n}$$
$$=\sum_{k\geq 0}\binom{2n}{k,k,2n-2k}-\sum_{k\geq 0}\binom{2n}{k,k+1,2n-2k-1}=|TPe_{2n}|.$$
Can it be made bijective?
Yes. All the manipulations with binomial coefficients have obvious bijective interpretations in terms of lattice paths, so one would have to do the same for the Jacobi-Trudi formula mentioned above. This is classical for the case of Catalan numbers, but it can be generalized for any partition using a similar reflection trick. I have been lazy and didn't work out the final bijection for you, but at least this is a start. I am currently writing a note about such bijections between SYT's with bounded parts, so if no one has explained it by then I'll make sure to update this answer.