Question

How many different numbers of six digits (without repetition of digit) can be formed from the digits 3, 1, 7, 0, 9, 5?
(i) How many of them will have 0 in the unit place?
(ii) How many of them are divisible by 5?
(iii) How many of them are not divisible by 5?

Answer 1

3, 1, 7, 0, 9, 5, i.e. 6 digits.
The total numbers of 6 digit numbers
6! - 5! = 720-120 = 600.
(i) 5! = 120 having 0 at the end.
(ii) Divisible by 5.
They will have zero in the last place and hence the remaining 5 can be arranged in 5! = 120 ways.
They may have 5 in the last place and as above we will have 5! = 120 ways. These' will also include numbers which will have zero in the first place. Therefore the numbers having zero in 1st and 5 in last place will be 4!
Therefore 6 digit numbers having 5 in the end will be
5! -4! = 120 - 2.4 = 96.
Therefore the total number of 6 digit numbers divisible by 5 is
120 + 96 = 216.
(iii) Not divisible by 5.
= Total - (divisible by 5)
=600 - 216 = 384.