Let T = {1000,1001...,9999}. How many numbers have at least one digit that is 0... Let T = {1000,1001...,9999}. How many numbers have at least one digit that is 0, at least one that is 1 and at least one that is 2. For example, 1072 and 2101 are two such numbers.
I have no idea to solve such probability question, I should use Permutation or Combination? why the answer is 150.
Thanks you guys.

Let's break up the different cases, shall we?


*

*If the leading digit $d$ is not one of the special digits, that leaves us with $d \in \left\{3,4,5,6,7,8,9\right\}$. So we have $7$ different options for the leading digit. After that, there are $3! = 6$ different options for our special digits — $\left\{(d,0,1,2),(d,0,2,1), (d,1,0,2),(d,1,2,0),(d,2,0,1),(d,2,1,0)\right\}$.
So the total number of combinations is $7 \cdot 3! = 42$.

*If the leading digit is a special digit $s$, it has to be either $1$ or $2$. It can't be $0$, as that would bring the number below $1000$, which is not allowed.  


*

*If the free digit $f$ is not equal to one of the remaining special digits, there are $3!$ ways of placing those digits, with $8$ options for the free digit — $f \in \left\{s,3,4,5,6,7,8,9\right\}$. This gives us $3! \cdot 8 = 48$ different options.  

*If $f$ is equal to one of the remaining special digits, there are $3$ ways of placing the other remaining digit and there are of course $2$ special digits that $f$ can be equal to, giving us $2\cdot3=6$ options.  


As noted earlier, there are $2$ possible leading special digits, giving us $2\cdot(48+6) = 108$ options for a leading special digit.
Adding these two cases gives us $42 + 108 = 150$, which is the answer you had.
