The question is:
Of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 90
(B) 82
(C) 80
(D) 45
(E) 36

The answer is C and I got that by thinking of what the possible combinations can be and summing them. If you look on page 51 of the Official Guide for the answer explanation, it has a way to solve the problem. However, I am not sure why this method is correct. I understand that for three-digit integers, when one of the pairs (tens and ones, hundreds and tens, hundreds and ones) is the same, there are 9 options for the third digit to be different from the other two. Where I am lost is that, since the question asks for integers greater than 700, say if you are given _99, then you only have 3 option for the third number: 7, 8, or 9. You don't have 9 options. However, you have many options as to which two numbers the given same tens and ones can be (_11, _22, etc.) Therefore I really don't understand this approach that the book is using to derive the answer. Although their answer comes to the same as mine.

I am lost... am I misinterpreting?

You aren't misinterpreting; you're just approaching the problem from a different angle. Your '3 options' observation is perfectly correct; this observation is implicit in the OG solution table, in the fact that there are exactly 3 columns in the table (corresponding to your '3 options' in each case).

There is a major flaw in their solution, by the way: the heading of the table ('Number of digits available...') only applies to the BOTTOM 2 ROWS. For the top row ('tens and ones'), the heading should be modified to say 'Number of digits available for the two selected digits,' because in that row it's the tens and ones digits (not the hundreds digit) that change.

Is there better simpler method to solve this problem ? I thought the method explained in OG is time consuming. Please advise. Thank you.

We can no longer discuss OG questions or post new OG questions, unfortunately - the above question pre-dates the legal request from GMAC so we can leave the archive up but we cannot discuss them any longer. Sorry :(

_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT