I Used choice Method in which first place has has Eight choices (Cannot be either one or zero) second Place has two choice and third place has nine choice removing zero = 144
but the correct answer is 152 what went wrong
When you chose 1 for second digit, 0 is still permissible to be used in the third digit. Say 510 is a valid code. But you have missed it because you eliminated the 0 for third digit for both cases of second digit. So better to break the cases of when second digit is 1 vs 0.
- 8 Possibilities for first digit (excluding 0 and 1)
- 1 possibility for second digit (1 only)
- 10 possibilities for third digit (all digits)
8 x 1 x 10 = 80
- 8 Possibilities for first digit (excluding 0 and 1)
- 1 possibility for second digit (0 only)
- 9 possibilities for third digit (excluding 0)
8 x 1 x 9 = 72
Now add the possibilities from the split cases to get the total number of possibilities.
Thanks 