Probability that a random k-digit number contains at least one 0, 1, and 2

Let $\displaystyle k\geq 3$ be any given integer. What is the probability that a random *k*-digit number will have at least one 0, at least one 1, and at least one 2?

I'm thinking that this could be answered via the principle of inclusion-exclusion, but I could use some clarification (or possible correction). Here's what I've got so far.

Assuming that leading zeroes are not allowed,

# of k-digits containing no 0’s: $\displaystyle 9^k$

# of k-digits containing no 1’s: $\displaystyle 8×9^{k-1}$

# of k-digits containing no 2’s: $\displaystyle 8×9^{k-1}$

# of k-digits containing no 0’s and no 1’s: $\displaystyle 8^k$

# of k-digits containing no 0’s and no 2’s: $\displaystyle 8^k$

# of k-digits containing no 1’s and no 2’s: $\displaystyle 7×8^{k-1}$

# of k-digits containing no 0’s, 1’s or 2’s: $\displaystyle 7^k$

I then applied the following:

$\displaystyle |A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|$

This would obtain the complement set of what I'm looking for, so this would be subtracted from $\displaystyle 9×10^{k-1}$.

Am I on the right track?