Question

How many five digit numbers divisible by $3$ can be formed using the digits $0,1,2,3,4,7$ and $8$ if, each digit is to be used atmost one.

Answer 1

Given digit 0,1,2,3,4,7,8

To be divisible by '3' sum should be divisible by '3'

so digit can be

0,1,2,4,8 ___ (1)

0,1,3,4,7 ___ (2)

1,2,3,4,8 ___ (3)

1,2,3,7,8 ___ (4)

Casess for (1) = $4\times 4!=96$

Cases for (2) = $4\times 4!=96$

Cases for (3) = $5!=120$

Cases for (4) = $5!=120$

432 Cases