How many three-digit numbers, which have an even tens digit and no repeated digits, can be formed using the elements of the set containing the numbers five, eight, nine, and two?
We’re not going to try and list out all possible three-digit numbers. Instead, we’re going to recall the product rule for counting, called the product rule because it involves some multiplication. It tells us that, to find the total number of outcomes from two or more events, we multiply the number of outcomes for each event together. So what are our events here? Well, we’re looking for a three-digit number, so each event is the digit we choose. There is a little bit of a restriction on this number. It must have an even tens digit. This tells us the tens digit can only be eight or two. So we’ll consider that digit first.
There are two digits that we can choose for the tens digit. So two is the number of ways of choosing that digit. We’ll now consider any other digit in the number, perhaps the hundreds digit next. We have chosen one number out of our set, meaning there are three numbers left to choose from. We know that there are only three numbers left to choose from because we don’t want there to be any repeated digits. So we say that the number of ways of choosing the hundreds digit in our three-digit number is three.
There is one digit left to choose. That’s the units digit. We’ve already taken two elements out of our set, meaning there are only two elements left to choose from, since we’re not repeating any digits. And so we found the number of outcomes for each event. That was the number of ways of choosing a tens digit, the number of ways of choosing a hundreds digit, and the number of ways of choosing a units digit.
The product rule says that to find the total number of three-digit numbers, we multiply these together. That’s two times three times two which is equal to 12. And so we see that there are 12 three-digit numbers which have an even tens digit and no repeated digits using elements of our set.