### Video Transcript

How many two-digit numbers, which
end with the digit two and have no repeated digits, can be formed using the elements
of the set three, one, two?

There are two ways we can answer
this question. The first is called systematic
listing. Since we have a very small set of
numbers from which to choose from, we know the outcomes for an event, that is, the
number of two-digit numbers we can make, can be listed or organized in a systematic
way. So, let’s see what that would look
like. We need our number to end with the
digit two. And so, this means its first
digit. Remember, we’re only interested in
a two-digit number. And the digits aren’t repeated can
either be three or one. And so, the numbers that we can
make are either 32 or 12. There are two ways to make
two-digit numbers which end with the digit two and have no repeated digit from our
set.

Now, an alternative method we
could’ve used is the product rule for counting. This can be nicer when we’re
dealing with a larger number of events. It says that to find the total
number of outcomes for two or more events, we multiply the number of outcomes for
each event together. In this case, our events are the
numbers we choose. So, we begin by looking at our
restriction: our two-digit number must end with the digit two. And so, there’s actually only one
way of choosing the second digit from the elements of our set.

Then, we look at the other digit in
our two-digit number. Remember, there are no repeated
digits and we have three numbers in our set. This means once we’ve chosen that
second digit, there are only two numbers left to choose from for the first
digit. The product rule says that we need
to multiply these to find the total number of outcomes. That’s one times two, which once
again gives us two.

There are two two-digit numbers
which end with the digit two that can be made using elements of the set containing
the numbers three, one, and two.