Given $10$ digits, where each digit can be an integer from $0$ to $9$, how can I determine the number of ways to arrange the numbers so that two odds are not adjacent?
Repetition of digits is not allowed.
So far, I have figured out the total number of possibilities: $$10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 10!$$
Then I had planned to subtract the number of bad possibilities from the total number of possibilities.$$10! - X$$ Where $X$ is all the bad possibilities, which means $X$ is all the possibilities where two odds could be next to each other in the $10$ digits.
I know that for each number, $5$ odds can be selected, how can I use this information to figure out the answer to the question?