Of the three-digit integers greater than 200, how many have two digits that are equal to each other and the remaining digit different from the other two?  Choices:
A) 90  B) 82  C) 80  D) 45  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; 200.   Answer:(36)
<result>E</result>
Of the three-digit integers greater than 200, how many have two digits that are equal to each other and the remaining digit different from the other two?