gclub tough one

[bblast]

How many five-digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits?

15
96
120
181
216

oa-E

[Frankenstein]

Hi,
For the 5-digit number to be divisible by 3, the sum of 5 digits has to be a multiple of 3. The digits of the number can be either 0,1,2,4,5 or 1,2,3,4,5 for the sum to be multiple of 3.
Case(1) : 0,1,2,4,5
Lets the number be of the form abcde.
a->1,2,4 or 5 -> 4ways
b->any of the other 5 except a -> 4ways
c->3
d->2
e->1
So, number of ways is 4.4.3.2.1 = 96

Case(2) : 1,2,3,4,5
Number of ways - >5! =120

Therefore, total no. of ways = 96+120 = 216

Cheers!