IMO C

suppose digits are d1,d2,d3,d4,d5
we 0,1,2,.......9 total 10 digits
for d1 we have 9 choices as 0 cannot take 5th dig place so 9 ways
for d2 we again have 9 choices as 0 is included now and d1 is excluded so 9 ways
for d3 again we have 9 choices as d1 is included and d2 is excluded so 9 ways
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like wise 9 choices for each other places too
total ways=9*9*9*9*9=9^5

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!

@thephoenix and @Brian, Thank a ton for the wonderful explanations

Whats the likelihood that a person in the 45-46 range in Quant would get a problem like this one in the actual test? Isn't this more of 49-51 level problem?

Hey, beatthegmatinsept (15 days left!):

To score 45-46, you'll almost definitely see problems in the 49-51 range...they have to test your "upper limit" with harder questions, so you'll see pretty much everything they have.

As far as permutations/combinations problems go, I wouldn't consider this one to be "devastatingly hard". I hate giving difficulty estimates, so take this as just a guess, but I'd probably say that this is closer to a 40-scaled-score problem than a 50. If P/C problems aren't your strength, it's obviously harder for you, but for those who have put a few hours into combinatorics I'd say that this is maybe average-level for combinatorics.

I've talked to at least one student whose FIRST question was a permutations/combinations problem, so this is definitely in the realm of what you could see regardless of how you're doing. But keep in mind that I don't think it's a problem that requires any true specific knowledge of permutations. If your only mode of thinking is "how many options are available for each digit" you can get there without specific knowledge, and if you haven't even thought that far you could use the answer choices as a guide ("well...it looks like I need to multiply with at least some 9s...how would I even start to do that?") to determine what kind of process to use.

Thanks Brian. When I go through the solution, it makes complete sense, but when I first read the problem, I almost had a brain freeze. I guess P&C just freaks me out, hence the reaction.
I am currently in the 44-46 range in Quant in most of my practice tests,maybe I should focus on getting some exposure to some P&C problems the next 10 days.