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?