Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?	Choose the correct choice:
A) 47
B) 60
C) 92
D) 95
E) 96
Three digit number can have only following 3 patterns:   A. all digits are distinct;   B. all three digits are alike;   C. two digits are alike and third digit is different.   We need to calculate C. C = Total - A - B   Total numbers from 660 to 999 = 339 (3-digit numbers greater than 660);   A. all digits are distinct = 3*8+3*9*8= 
<gadget id="calculator">3*8+3*9*8</gadget>
<output>240</output>
240. If first digit is 6 then second digit can take only 3 values (7, 8, or 9) and third digit can take 8 values. If first digit is 7, 8, or 9 (3 values) then second and third digits can take 9 and 8 values respectively;   B. all three digits are alike= 4 (666, 777, 888, 999).   So, 339-240-4=95.   Answer: 95
<result>D</result>
Of the three-digit integers greater than 660, how many have two digits that are equal to each other and the remaining digit different from the other two?