# Calculate the expected value of the number of different digits

Question. $$n$$ is a $$m$$-digit number. ($$m$$ can start with zero.) Let $$P(n)$$ the number of different digits in $$n$$. What is the expected value of $$P(n)$$? (For example, $$P(12341234)=4$$.)

My approach.

Let X:={$$(n, q)$$|$$n$$ is $$m$$-digit number, $$q$$ is the set of different digits in $$n$$}. (For example, $$(123123, \left\{ 1, 2, 3\right\}) \in X$$)

Let's use double counting.

It is clear that $$|X| = 10^m$$.

I think when I calculate the sum of $$|q|$$ in all of the elements of $$X$$, then I can get the answer.

What should I do to calculate it? (Maybe I need the inclusion–exclusion principle.)

Use the linearity of expectation. What is the chance that $$1$$ is present? That is the expected value of an indicator variable.