Initially thought that it would be
$$(9C1)\cdot(8C1)\cdot(7C1)\cdot(6C1)\cdot(5C1)\cdot(4C1)\cdot(2C1)\cdot(1C1)$$
but I soon realised that it was restricting its order and thus creating additional combinations. I’ve been trying it for an hour now but I still can’t seem to come up with an intuition behind this. ANY help would be greatly appreciated . Also I am sorry for not using the symbols properly. I’m new to maths stackexchange and I’m still struggling with mathjax. I’ll soon update the answer with apropriate symbols
EDIT: I realised I made a mistake in the answer. Here is the updated version
$$9 \cdot 7 \cdot 5 \cdot 3 = 945$$
Best Answer
Hint. The answer can be written as $$9\cdot \frac{\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}}{4!}\quad \mbox{or}\quad \frac{\binom{9}{2}\binom{7}{2}\binom{5}{2}\binom{3}{2}}{4!}\cdot 1.$$ Can you picture what is happening?