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

Let 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:
# of k-digits containing no 1’s:
# of k-digits containing no 2’s:
# of k-digits containing no 0’s and no 1’s:
# of k-digits containing no 0’s and no 2’s:
# of k-digits containing no 1’s and no 2’s:
# of k-digits containing no 0’s, 1’s or 2’s:

I then applied the following:

This would obtain the complement set of what I'm looking for, so this would be subtracted from .

Am I on the right track?

You are.

Thanks for confirming that.

I currently have the answer written out as follows:

This probably isn't the cleanest way to show the answer. Is there a more condensed version I could use?

There is probably no shorter formula for this case.