This is a question that came to mind and I was trying multiple ways from using combinatorial methods, graphical approaches, etc. The question I have is, how can I find the product of all the odd or even numbers less than some positive integer $n$.
For example:
If $n=6$, then the product of all odd numbers less than 6 would be $15$ and all even numbers would be $8$. But this gets impossible after you reach larger integers.
Best Answer
Case for Odd Product:
Since you want the product less than $n$, you are looking for the product of, $$ 1\cdot 3\cdot 5\cdot 7\cdot 9\cdot \ldots \cdot (n-1) $$ Now, notice that you don't want the even numbers, so just divide by them, $$\frac {1\cdot 2\cdot 3\cdot 4\cdot 5\cdot \ldots \cdot n} {2\cdot 4\cdot 6\cdot 8\cdot 10\cdot \ldots \cdot n} = \frac {1\cdot 2\cdot 3\cdot 4\cdot 5\cdot \ldots \cdot n} {(2\cdot 1) \cdot (2\cdot 2) \cdot (2\cdot3) \cdot (2\cdot 4) \cdot (2\cdot 5)\cdot \ldots \cdot \left(2\cdot \frac n2\right)} = \displaystyle\frac{n!}{2^{\frac n2}\left(\frac n2\right)!} $$
Now take a similar approach to find even products.