What is the probability that a ten digit number that is a number chosen at random between 1,000,000,000 and 9,999,999,999 inclusive will have ten different digits?

Answer 1

The number of such "numbers"
= 9*9*8*7*6*5*4*3*2*1
we can use 9 different digits at the front, once that is filled, 9 are left to
pick from, then 8, etc
= 9*9!
number of possible 10 digit numers = 9*10^9
once the first digit is filled, there are 10 for each remaining,
digits may be repeated, including the 0

so prob(your event) = 9*9!/(9*10^9)
= 00036288 (on my calculator)