So I'm trying to construct a 6-digit combination number where the digits can be from 0 to 9 and it must have at least one odd and one even number in the combination and there are allowed to be repeats of a number.
My intuition tells me that from the $10^6$ possible combinations the easiest way to work this out would be to take away the combinations where they are all even or all odd.
So $10^6 - 5^6 - 5^6 = 968750$
Is this correct or am I missing something?
Yes, you are correct.
Here is a slower method just to verify things where we let $i$ be the number of odd digits.
$$\sum_{i=1}^5 \binom{6}{i}5^i \cdot 5^{6-i}=5^6 \sum_{i=1}^5\binom{6}{i}=5^6(2^6-2)=10^6-2\cdot 5^6$$