The first move is necessary for the shortest checkmate. Then, the three first uneven moves have 4 possibilities for black (the other uneven ones have 2), and each even move (except the last one) has 2 possibilities for white.
So... there are 4096, if I counted right.
... Then How Many Possibilities Should There Be?
This is obviously not dual-free. However the riddle at hand is:
"How many possibilities are there from move 1 to the shortest checkmating move at the end?"
You have to count the parallel-universes here. This puzzle-form is 1 of those universes. It 1 is called "Peekaboo":