How many positive three-digit integers have the hundreds digit equal to 7 and the units digit equal to 1?

This is an example of a permutation problem.

Since there are three digits, the total number of possibilities would be:

total number of possibilities = (amount of numbers that can take up hundreds place) * (amount of numbers that can take up tens place) * (amount of numbers that can take up ones place)

It is given that there is only one number that can take up the hundreds and ones place, 7 and 1 respectively. For the tens place, there are ten numbers that can take this place (0 to 9), therefore:

total number of possibilities = 1 * 10 * 1

total number of possibilities = 10

There are 10 positive three -digit integers.