How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
A
5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
10
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
15
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
20
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D 20 Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it. The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place. The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it. ∴ Required number of numbers = (1 x 5 x 4) = 20.