Rolling dice possible outcomes You roll $10$ dice. How many possible outcomes include the numbers $1$
  to $6$?


It's essentially asking how many outcomes there are for 1 die to be 1, another to be 2, another to be 3, another to be 4, another to be 5, and another to be 6.

I know that we have $6 \cdot 5\cdot 4\cdot 3\cdot 2\cdot 1 = 6!$ ways for 6 dice to contain 1-6, but I'm not sure about 10 dice. Thanks
Use Inclusion and Exclusion. All the ways to select at most 6 distinct faces, minus all the ways to select at most 5 of the 6, plus all the ways to select at most 4 four of the six, minus...

$$\dbinom 6 6~6^{10} - \dbinom 6 5~ 5^{10} + \dbinom 6 4~ 4^{10} - \dbinom 6 3~ 3^{10} + \dbinom 6 2~ 2^{10}-\dbinom 6 1 ~1^{10} + \dbinom 6 0 ~0^{10}$$
Based on the comments below, I try a different approach. I consider $1-$the number of cases where all numbers are not there. 

Choose 5 out of 6 numbers. Each dice can be any number = $5^{10}$ 

So, the total number of ways is $6^{10} - 6*5^{10}$