sondenso wrote:

of the three-digit integers greater than 700, how many have two digits that are equal to each other and the remaining digit diffferent from the other two?

A. 90

B. 82

C. 80

D. 45

E. 36

7_ _ ( 700 < Total number of digits <=799 is 99)

8_ _ ( 800 <= Total number of digits <=899 is 100 )

9_ _ ( 900 <= Total number of digits <=999 is 100 )

Consider 7 _ _

Lets calculate number of three digit numbers greater than 700 which have all three DIFFERENT numbers.

7_ _

second digit can take ONLY 9 digits. (becoz we need to exclude 7)

third digit can take ONLY 8 digits. (bcoz we need to exclude 7 and the above number that we already selected.)

So total = 9 * 8 = 72

Now TOTAL - ALL Different numbers = numbers which have same digits in it

99 - 72 =27

But wait this includes 777, so 27 - 1 = 26

Similarly for 8 _ _,

100 - 72 = 28

But wait this includes 888, so 28 - 1 = 27

Similarly for 9 _ _,

100 - 72 = 28

But wait this includes 999, so 28 - 1 = 27

Total = 26 + 27 +27 =80

C