# Finding product of odd/even integers

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.

• You can get expressions in terms of factorials. But you may not find that satisfactory, factorials are also hard to compute. – André Nicolas Jan 5 '16 at 0:59
• You may be interested in Double Factorial notation. $5!! = 5\cdot 3\cdot 1$ whereas $6!!=6\cdot 4\cdot 2$. – JMoravitz Jan 5 '16 at 1:01
• Hint: if $n=2m$ were even, say, then the product of the even numbers less than or equal to $n$ would be $2^m\,m!$. – lulu Jan 5 '16 at 1:01
• Laborious, yes, impossible no. – Robert Soupe Jan 5 '16 at 4:18

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, $$\frac {1\cdot 2\cdot 3\cdot 4\cdot 5\cdot \ldots \cdot n} {2\cdot 4\cdot 6\cdot 8\cdot 10\cdot \ldots \cdot n} = \frac {1\cdot 2\cdot 3\cdot 4\cdot 5\cdot \ldots \cdot n} {(2\cdot 1) \cdot (2\cdot 2) \cdot (2\cdot3) \cdot (2\cdot 4) \cdot (2\cdot 5)\cdot \ldots \cdot \left(2\cdot \frac n2\right)} = \displaystyle\frac{n!}{2^{\frac n2}\left(\frac n2\right)!}$$
If $n$ is even you get that the even product is $2^{\frac{n}{2}}\left(\frac{n}{2}\right)!$ The whole product is $n!$, so the odd product is $\frac{n!}{2^{\frac{n}{2}}\left(\frac{n}{2}\right)!}$ If $n$ is odd just change $n$ with $n-1$ in the even product formula.