Hey Sanalnnair,
Great question - this is a good example of a permutations/combinations question in which you need to determine how many options are available for each space. Let's break down the number of options for each:
First Space (or the Ten-Thousands digit)
9 options (1, 2, 3, 4, 5, 6, 7, 8, 9 - if it were 0, that wouldn't be a five-digit number)
Second Space (the Thousands digit)
9 options (anything but the first one, as we can't have any consecutive digits, but 0 is now a possibility)
Third Space (the Hundreds digit)
9 options (anything but the second one, again to avoid consecutive digits)
Fourth Space (the Tens digit)
9 options (anything but the third one, again to avoid consecutive digits)
Fifth Space (the Units digit)
9 options (anything but the fourth one, again to avoid consecutive digits)
Because there are 9 options for each space, we'd multiply 9 * 9 * 9 * 9 * 9 to get 9^5.
Why do we multiply? Let's try this with three spots and three options, A, B, and C:
The first spot lets us have 3 choices, A, B, or C
For each one that we pick, there are two options left:
Pick A, you could then go B or C
Pick B, you could then go A or C
Pick C, you could then go A or B
Because each first option has its own set of second options, you'd multiply those together, and so on.
The biggest key for most combinatiorics problems is simply determining how many options you have for each space, and knowing that you do a lot of multiplication in these problems!
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep
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