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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 ${a+b \choose b}_q$.

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 by Melczer, Panova, and Pemantle.

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 quantity you are interested in 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 ${a+b \choose b}_q$.

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 ${a+b \choose b}_q$.

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 quantity you are interested in is obtained by specializing at $q=-1$. Depending on the parity, one either gets 0 or the middle (and largest) coefficient of the polynomial ${a+b \choose b}_q$.

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 quantity you are interested in 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 ${a+b \choose b}_q$.