There is a phone number with $10$ digits. What's the probability that each digit occurs exactly once?

Question

There is a phone number with $10$ digits. What's the probability that each digit occurs exactly once?

This is a task of an old exam and I'm curious if my solution is correct.

So we have a phone number made up of $10$ digits. Any number between $0$ and $9$ can occur as a digit. Thus in total we have $10^{10}$ combinations. But we want that each digit occurs exactly once. Since we have length $10$ with $10$ different numbers, there are $10!$ different possibilities of arranging those numbers (where each number occurs exactly once).

What we need to do is $10^{10}-10!$

Thus the probability that each digit occurs exactly once in a phone number of length $10$ is $$\frac{1}{10^{10}-10!}$$

?

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By the way, it took me about $15$ minutes to get to that solution and in the exam I got like $5$ minutes for one task.. :p

I hope it's correct at least?

Answer 1

What you need is the possibilities for the combination of interest (here: all different digits), divided by total possibilities. You calculated the components correctly but not how to put them into a probability. So

$$\frac{10!}{10^{10}} = \frac{9!}{10^{9}} \approx 0.000363$$

which is significantly larger than your result.