Funky Factorials

For any double factorial, here are the cases: $n!! = \begin{cases} n \times (n-2) \times \ldots \times 5 \times 3 \times 1 , & n \text{ odd, } \\ n \times (n-2) \times \ldots \times 6 \times 4 \times 2 , & n \text{ even, } \\ 1, & n = 0, - 1 \\ \end{cases}$

This question, belongs to the first case, hence $3!! = 3\times (3-2)\\ = 3\times 1 \\ = \boxed{3}$

Don't get confused between $3!!$ and $\left( 3! \right) !$

$!!$ denotes double factorial and it is defined as

$n\left( n-2 \right) \left( n-4 \right) \dots 4\cdot 2$

if $n$ is even and

$n\left( n-2 \right) \left( n-4 \right) \dots 3\cdot 1$ if $n$ is odd.

Therefore $3!! = 3$