I was doing a bit of research into factorials, and found both hyperfactorials (denoted by an $H$) and alternating factorials (denoted by an $AF$). Hopefully this answer fulfills your need.
$AF(\sqrt{9} + 1) * H(2 + 0)$
First we can take the hyperfactorial of $H(2 + 0)$.
$AF(\sqrt{9} + 1) * 4$
We'll then solve for the other pair of brackets.
$AF(4) * 4$
Now we'll take the alternating factorial.
$19 * 4$
And then some basic multiplication to get:
$76$