### Instruction: 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?### Response: To find the probability of an event, I need to divide the number of favorable outcomes by the number of possible outcomes.
The possible outcomes are the ways to choose 5 integers from between 20 and 69, inclusive.
That's 50 numbers, so I can use the combination formula to find the number of ways to choose 5 of them: 50 choose 5, or 50!/(5!(50-5)!) = 2,118,760.
The favorable outcomes are the ways to choose 5 integers that each have a different tens digit.
That means I need to pick one number from each of the 5 groups: 20-29, 30-39, 40-49, 50-59, and 60-69.
There are 10 numbers in each group, so for each group I have 10 choices.
The total number of ways to pick one from each group is 10 times 10 times 10 times 10 times 10, or 10^5 = 100,000.
So the probability of picking 5 integers that each have a different tens digit is 100,000/2,118,760, which I can simplify by dividing both numerator and denominator by 40 to get 2,500/52,969.
This fraction is already in lowest terms, so I can leave it as my final answer, or I can convert it to a decimal or a percentage by dividing 2,500 by 52,969.