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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 factor.

$19 * 4$

And then some basic multiplication to get:

$76$

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:

$76$

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.

$AF(\sqrt{9} + 1) * H(2 + 0) = 76$

First we can take the hyperfactorial of $H(2 + 0)$.

$AF(\sqrt{9} + 1) * 4 = 76$

We'll then solve for the other pair of brackets.

$AF(4) * 4 = 76$

Now we'll take the alternating factor.

$19 * 4 = 76$

And then some basic multiplication to get:

$76 = 76$

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 factor.

$19 * 4$

And then some basic multiplication to get:

$76$

I was doing a bit of research into factorials, and found these little gems called superfactorials…
 

$superfactorial(\sqrt{9}) + 2 + 1 + 0! = 76$

Superfactorials are the multiplied value of every factorial number before it.

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.

$AF(\sqrt{9} + 1) * H(2 + 0) = 76$

First we can take the hyperfactorial of $H(2 + 0)$.

$AF(\sqrt{9} + 1) * 4 = 76$

We'll then solve for the other pair of brackets.

$AF(4) * 4 = 76$

Now we'll take the alternating factor.

$19 * 4 = 76$

And then some basic multiplication to get:

$76 = 76$