[Guest]

N is a 15-digit base ten positive integer with no leading zero.

Determine the probability that N contains each of the digits from 0 to 9 at least once, but at most twice.

[Guest]

Number of ways in which the 15 digit number can be written under the rules mentioned can be calculated by taking 0 to 9 all digits and choosing 5 digits out of 10 to complete the number and then rearrange the digits

Now, we consider two cases, one where 0 is one of the chosen 5 extra digits and one where 0 is not chosen

case 1) where 0 is not chosen

5 digits can then be chosen in 9C5 ways

Now we need to arrange them such that 0 is not the leading digit. This can be done then in 9C5 x 14! x 14 ways

Case 2) where 0 is one of the chosen 5 digits

then other 4 digits can be chosen in 9C4 ways

Now these 15 digits can be arranged in

9C4 x 13! x 13 x 14 = 9C4 x 14! x 13 ways

Total number of ways in which the 15 digit number can be written as per rules mentioned then is 9C5 x 14! x 27

Total number of 15 digit numbers = 9 x 10^14

Probability of number being written as mentioned then is: 9C5 x 14! x 27 / (9 x 10^14) = 0.329

Almost 1/3rd of the numbers!!