shrive555
A popular website requires users to create a password consisting of digits only. If no digit may be repeated and each password must be at least 9 digits long, how many passwords are possible?
9! + 10!
2 x 10!
9! x 10!
19!
20!
i got it right with slot method, but wasn't sure.

shrive555
why 9! is multiplied with 10 ?
thanks
Note that we are told that no digit may be repeated.There are total of 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
As password must be
at least 9 digits long then it can consist of either 9 or all 10 digits.
If password consists of 9 digits then # of passwords possible is \(C^9_{10}*9!=10*9!=10!\): \(C^9_{10}\) - # of ways to choose 9 digits which will be used in password and \(9!\) arranging these digits (or in another way \(P^9_{10}=10!\) - choosing 9 digits out of 10 when the order matters);
If password consists of all 10 digits then # of passwords possible is simply \(10!\);
So total # of passwords possible is \(10!+10!=2*10!\).
Answer: B.