govind_raj_76 wrote:How many numbers from 1 to 1000, both inclusive, have digits repeated ?

A) 9

B) 81

C) 262

D) 648

E) 738

Please explain the concept behind this and how to solve this problem ?

Thanks,

i am not too sure in my answer, i got C-262
i calculate the number of integers consisting of different integers and substract this number from 1000

1,2,3,4,5,6,7,8,9 total-9 numbers

two digit numbers, the option for the first digit- any digit but 0-9 options for the second digit 9 digit as one is already used
9*9=81

three digit numbers
9*9*8=648 numbers

in total=9+81+648=738
1000-738=262