The number of numbers lying between 99 and 1000 which can be formed using the digits 2, 3, 7, 0, 6, 8 when no digit is being repeated is
Digits are 2, 3, 7, 0, 6, 8 i.e. six in all.
We have to form number between 99 and 1000. Clearly they will be of three digits and their number will be
6P3=6×5×4=120
Out of these we have to exclude those numbers of 3 digits which have zero in the first place.
When first place is filled by 0, remaining can be arranged in 5P2 ways.
Their number is 5P2=5×4=20
∴ The required number is 120-20=100