Of the three-digit integers greater than 600, how many have two digits that are equal to each other and the remaining digit different from the other two? Choose the correct choice from the following choices.
A) 90
B) 82
C) 80
D) 75
E) 36
Let the three digit number be represented as X Y Z.   There are 3 cases:   Case I.[ X=Y ]Z is not equal to XY :XXZorYYZ   X can be either 7, 8 or 9, so digit at X can be chosen in 3 ways   After X is chosen, Y can be chosen in 1 way   After XY are chosen, Z can be chosen in 9 ways   Thus, possible No of digits = (3 ways) * (1 way) * (9 ways) = 27 ....(1)   [example numbers: 774,779,882,993 etc]   Case II.[ X=Z ]Y is not equal to XZ:XYXorZYZ   X can be either 7, 8 or 9, so digit at X can be chosen in 3 ways   After X is chosen, Z can be chosen in 1 way   After XZ are chosen, Z can be chosen in 9 ways   Thus, possible No of digits = (3 ways) * (9 ways) * (1 way) = 27 ....(2)   [example numbers: 747,797,828,939 etc]   Case III.[ Y =Z ]X is not equal to YZ :XYYorXZZ   X can be either 7, 8 or 9, so digit at X can be chosen in 3 ways   After X is chosen, Y can be chosen in 9 ways   After Y is chosen, Z can have 1 way   Thus, possible No of digits = (3 ways) * (1 way) * (9 ways) = 27 ....(3)   [example numbers: 744,799,822,933 etc]   Therefore, total numbers of possible digits [sum of Case (1), (2)(3) above] = 27 + 27 + 27 - 1= 
<gadget id="calculator">27 + 27 + 27 - 1</gadget>
<output>80</output>
80   One digit is subtracted from total number of possible digits to eliminate one possibility of XYZ= 700 to satisfy the condition that digit &gt; 600.   Answer:(75)
<result>D</result>
Of the three-digit integers greater than 600, how many have two digits that are equal to each other and the remaining digit different from the other two?