WEBVTT
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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?
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We’re not going to try and list out all possible three-digit numbers.
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Instead, we’re going to recall the product rule for counting, called the product rule because it involves some multiplication.
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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.
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So what are our events here?
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Well, we’re looking for a three-digit number, so each event is the digit we choose.
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There is a little bit of a restriction on this number.
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It must have an even tens digit.
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This tells us the tens digit can only be eight or two.
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So we’ll consider that digit first.
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There are two digits that we can choose for the tens digit.
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So two is the number of ways of choosing that digit.
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We’ll now consider any other digit in the number, perhaps the hundreds digit next.
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We have chosen one number out of our set, meaning there are three numbers left to choose from.
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We know that there are only three numbers left to choose from because we don’t want there to be any repeated digits.
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So we say that the number of ways of choosing the hundreds digit in our three-digit number is three.
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There is one digit left to choose.
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That’s the units digit.
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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.
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And so we found the number of outcomes for each event.
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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.
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The product rule says that to find the total number of three-digit numbers, we multiply these together.
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That’s two times three times two which is equal to 12.
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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.