What is the probability that a randomly chosen 10-digit number is a valid ID?

Question

What is the probability that a randomly chosen 10-digit number is a valid ID subject to the following rules: Each ID is a permutation of digits of 0-9. The digits 3 and 7 must occur as the substring 37. The digits 2 and 4 must occur as the substring 42. The digits 5, 6, 8 must occur as the substring 856.

I know that when considering these rules there are a total of 6 different numbers that can be chosen from, 42, 73, 856, 0, 1, 9. I am just not sure how to figure out the probability of what is being asked. Thanks in advance.

Answer 1

You're well on the way!

However, you don't have just $6$ possible valid ID's, but rather you have $6!$ valid 10-digit ID's, since that is the number of permutations for the $6$ substrings $42$, $73$, $586$, $0$, $1$, $9$.

Also, this is out of $10^{10}$ possible 10-digit strings. So, the probability is ....

Answer 2

Guide:

A chosen $10$ digits number can have duplicates of digits., it need not be a permutation of the digits. Neither must it follow those states rules.

There are $10^{10}$ such options of a $10$-digits number, including those that are invalid ID.