1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# 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

A
102
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
120
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
100
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is C 100 Digits are 2, 3, 7, 0, 6, 8 i.e. six in all. We have to form a 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 the first place is filled by 0, remaining numbers can be arranged in 5P2 ways. Their number is 5P2=5×4=20 ∴ The required number is 120-20=100

Suggest Corrections
1
Join BYJU'S Learning Program
Related Videos
Factorising Numerator
QUANTITATIVE APTITUDE
Watch in App
Join BYJU'S Learning Program