Suppose 5 different integers are randomly chosen from between 20 and 69, inclusive. What is the probability that they each have a different tens digit?

Guest Jun 27, 2018

#1**+2 **

From 20 to 69, they are a total of 50 integers.

To choose 5 of them, we do: \(\binom{50}{5}\)

Now, we need to find out the number of successful outcomes.

We need an integer between 20 and 29

Then another one from 30 to 39, 40 to 49, 50 to 59, and 60 to 69.

There are 10 numbers in each category, and therefore \(10^5\) ways of doing this.

Your final answer is: \(\boxed{\frac{2500}{52969}}\)

I hope this helped,

Gavin

GYanggg Jun 27, 2018