Combinations with divisibility

Question

How many 4 digit numbers, which are not divisible by 5 can be formed using the digits 4,5,6,7 without repeating any digit?

I have tried by using this formula nPr my answer is 24? is it right?

score 1 · Answer 1 · answered Apr 24 '16 at 07:14

1

The total number of combinations is $4!=24$.

The number of combinations ending with $5$ is $3!=6$.

The number of combinations not ending with $5$ is $24-6=18$.

answered Apr 24 '16 at 07:14

barak manos

43,109

score 0 · Accepted Answer · answered Apr 24 '16 at 07:13

0

Not quite. If we permute the 4 digits we get 4! = 24 possibilities, but some of these possibilities have 5 in the units place i.e. are divisible by 5. I believe you should get 18

answered Apr 24 '16 at 07:13

kcborys

574

score 0 · Answer 3 · answered Apr 25 '16 at 00:24

Or to see it a third way. There are 3 possible choices for the last digit, a:4,6,and 7. There are 3 possible chosen for the third digit, b: the three that aren't a. There are 2 choices for the second digit, c: the two that aren't a or b. There is one chose for the first digit: whatever is left.

So there are $3*3*2*1=18$ possible ways to do this.

Combinations with divisibility

3 Answers3