In other words, ALL combinations which don't contain at least one of the number from 1-6 would count.
So for example...
5, 2, 3, 3, 4, 1, 5, 5, 3, 1 would be counted because there is no 6
Also
5, 3, 3, 3, 4, 1, 5, 5, 3, 1 would be counted as well because there is no 2 or 6
I am trying to figure out how many combinations/what percentage of the combinations out of 6^10, contain AT LEAST one missing number from 1-6.
How do I go about figuring this out?
The probability of not obtaining a specific result is $\left(\dfrac{5}{6}\right)^{\!10}$, and there are $6$ specific results.
But simply adding them together over counts the events of not obtaining 2 or more specific results.
The probability of not obtaining $k$ specified results is $\left(\dfrac{6-k}{6}\right)^{10}$, and there are $\dbinom{6}{k}$ ways of choosing $k$ specific results.
Now, can you use the Principle of Inclusion and Exclusion to put this together?