of the 3 - 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 ? Pick one:  A) 90  B) 82  C) 80  D) 45  E) 36

Rationale: 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 = 80 one digit is subtracted from total number of possible digits to eliminate one possibility of xyz = 700 to satisfy the condition that digit > 200 . answer : ( e )
Answer: E