A $(p,q)$-shuffle is a permutation $\sigma$ of the set $\{1, \dots, p, p+1, \dots,p+q\}$ such that $$\sigma(1)<\dots<\sigma(p)$$ and $$\sigma(p+1) < \dots<\sigma(p+q)\,.$$


It is known that the [number of $(p,q)$-shuffles](https://en.wikipedia.org/wiki/Riffle_shuffle_permutation) is ${p+q \choose p}$.


Looking at the case $p+q = 6$ and considering the parity of shuffles, I was surprised to find that


* $3$ of the ${6 \choose 1} = 6$ $(1,5)$-shuffles have even parity
* $9$ of the ${6 \choose 2} = 15$ $(2,4)$-shuffles have even parity
* $10$ of the ${6 \choose 3} = 20$ $(3,3)$-shuffles have even parity
* $9$ of the ${6 \choose 4} = 15$ $(4,2)$-shuffles have even parity
* $3$ of the ${6 \choose 5} = 6$ $(5,1)$-shuffles have even parity


It's not so difficult to believe that the number of even $(p,q)$-shuffles should equal the number of even $(q,p)$-shuffles. But I was very surprised to see that there are so many more even than odd $(2,4)$-shuffles.


Are any closed-form solutions known for the number of even $(p,q)$-shuffles? If not, is anything known about the asymptotic behaviour?



Parity of shuffle permutations
As Philippe suggests in the comments, it is well know that 
$$\sum\_{\sigma \text{ is a $(a,b)$-shuffle}} q^{\ell(\sigma)}={a+b \choose b}\_q$$
where the $q$-binomial coefficient (or Gaussian binomial coefficient) on the right side is defined by 
$${n \choose k}\_q=\frac{[n]\_q! }{[k]\_q! [n-k]\_q!}$$
where $[n]\_q!=[1]\_q [2]\_q \cdots [n]\_q$ and $[n]\_q=1+q+\cdots +q^{n-1}$.


Thus the difference between the number of even and odd shuffles is obtained by specializing at $q=-1$. This polynomial is palindromic, so, depending on the parity, one either gets 0 or the middle (and largest) coefficient of the polynomial ${a+b \choose b}\_q$.


Precise asymptotics for this coefficient (and others) are given in Theorem 1 of [this paper](https://arxiv.org/pdf/1805.08375.pdf) by Melczer, Panova, and Pemantle.


