Question

How many numbers formed with digits 0, 1, 2, 5, 7, 8 will fall between 13 and 1000 if digits can be repeated?

Answer 1

Case I:
2-digit numbers more than 13, less than 20, formed from the digits 0, 1, 2, 5, 7, 8
Number of such numbers = 3
Case II:
2-digit numbers more than 20 formed from 0, 1, 2, 5, 7, 8
Ten’s place digit is selected from 2, 5, 7, 8.
∴ Ten’s place digit can be selected in 4 ways.
Unit’s place digit is any one from 0, 1, 2, 5, 7, 8
∴ Unit’s place digit can be selected in 6 ways.
Using the multiplication principle, the number of such numbers (repetition allowed)
= 4 × 6 = 24
Case III:
3-digit numbers formed from 0, 1, 2, 5, 7, 8
100’s place digit is anyone from 1, 2, 5, 7, 8.
∴ 100’s place digit can be selected in 5 ways.
As digits can be repeated, the 10’s place and unit’s place digits are selected from 0, 1, 2, 5, 7, 8
∴ 10’s place and unit’s place digits can be selected in 6 ways each.
Using multiplication principle, the number of such numbers (repetition allowed) = 5 × 6 × 6 = 180
All cases are mutually exclusive and exhaustive.
∴ Required number = 3 + 24 + 180 = 207