Valeria thinks of a five-digit number that satisfies the following criteria. All of the digits are nonzero. The second and third digits are the same. The five-digit number is odd. How many different five-digit numbers could Valeria have thought of?
We can add some boxes for each digit of our five-digit number. We can call each digit a, b, c, d, and e, respectively. We’re told that none of the digits can be zero. That leaves us with a total of nine possible digits so we can choose from: one to nine.
However, there’s some more criteria. We’re told that the third digit c has to be the same as the second digit b. This means that whilst there is nine possible options for the digit named b, there is only one possible option for the third digit c.
We’re also told that the five-digit number is odd. That means it has to end in the numbers one, three, five, seven, and nine. So for the fifth digit, there are only five possible options.
So to recap, a and b, the first and second digits, have nine possible options. The third digit c has one possible option. The fourth digit d has nine possible options again. And the fifth and final digit e has five possible options.
The product rule for counting says that to find the total number of outcomes for two or more events — in this case, there are five so-called events and there the value of each digit in the number — we multiply the number of outcomes for each event together.
In this case, that would be nine multiplied by nine multiplied by one multiplied by nine multiplied by five. That gives us a total of 3645 possible outcomes. In this case, those different outcomes are five-digit numbers.
So there are 3645 unique five-digit numbers that Valeria could have thought of.