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Of the threedigit integers greater than 800
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02 Aug 2017, 09:05
With the given conditions, three types of numbers are possible
Type 1: AAB  First and Second Digits are same, third digit is different Type 2: ABA  First and Third Digits are same, Second digit is different Type 3: BAA  Second and third Digits are same, First digit is different
Type 1: First and Second Digit can be 8 and 9, Third Digit can be 0,1,2,3,4,5,6,7,8/9. (If first and second digit is 8, then third digit cannot be 8 and if first and second digit is 9, then third digit cannot be 9).
Therefore total no. of combinations for Type 1 = 2 *9 = 18
Type 2: First and Third Digit can be 8 and 9, Second Digit can be 0,1,2,3,4,5,6,7,8/9. (If first and third digit is 8, then Second digit cannot be 8 and if first and third digit is 9, then second digit cannot be 9).
Therefore total no. of combinations for Type 2 = 2 *9 = 18
Type 3: If First Digit is 8, then second and third digit can be 1,2,3,4,5,6,7,9. If First Digit is 9, then second and third digit can be 0,1,2,3,4,5,6,7,8.
Therefore total no. of combinations for Type 3 = 8+9 = 17
Total number of combinations = 18 + 18 + 17 = 53
Answer is c
