TITLE: What is the probability that a randomly chosen four-digit integer has distinct digits?
QUESTION [5 upvotes]: What is the probability that a randomly chosen four-digit integer has distinct digits?
first digit can't be 0 so it'd be between 9 numbers.
second digit can be 0 so it'd be between 9 numbers.
third would cant be either of the last two so between 8 numbers.
fourth would be between 7.
9*9*8*7/9000 = 0.504

REPLY [3 votes]: A slightly different way to do is just counting all the numbers with four distinct digits (including numbers starting by zero) and substracting the invalid numbers, that are a $10\%$ of this total, that is $$V=10\cdot 9\cdot 8\cdot 7-(10\cdot9\cdot 8\cdot 7)/10=9\cdot9\cdot8\cdot 7$$
Then if we call the event to get a number of four digits with all digits different $A$ (supposing that the probability to take a number is the same for all numbers) we have that
$$\Pr[A]=\frac{9\cdot9\cdot8\cdot 7}{10^3\cdot 9}=0.504$$