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by Cheese12 » Mon Oct 03, 2011 8:31 am
Of all the three-digit integers greater than 700, how many have two digits that are equal to each other and then remaining digit different from the other two ?

A) 90
B) 82
C) 80
D) 45
E) 36


OA:C

please tell how to solve this one... thanks!
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by shankar.ashwin » Mon Oct 03, 2011 8:36 am
There are 299 Numbers between 700 and 999.

No. of numbers where all are distinct = 3 * 9 * 8 = 216.

All 3 are same is 777,888 and 999.

So 2 equal and other diff = 299 - 216 - 3 = 80
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by GmatMathPro » Mon Oct 03, 2011 8:45 am
Focusing on 700's first, it must be of the form:

77X
7X7
7XX

where X is not equal to 7,we have 9 choices for X in each one (any of the ten digits except 7). 9+9+9=27. It would be the same for the 800s and 900s, so 27*3=81. However, this counts 700, and we only want numbers GREATER than 700, so subtract one to get 80.
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