A five digit integer divisible by 3 can be formed if:

1. 0 is exluded 1+2+3+4+5= 15, divisible by 3

and

2. 3 is excluded 0+1+2+4+5= 12, divisible by 3

1. 0 is excluded - 1,2,3,4,5
you can make a five digit number with distinct digits in 5! ways = 120

2. 3 is exluded - 0,1,2,4,5
First digit- 4 ways - 0 cannot be the first digit
Second digit- 4 ways
third - 3 ways
fourth - 2 ways
fifth -1 way
Thus, 4*4*3*2*1= 96

Total ways = 120+96 = 216

here's the oa:

The sum of digits of a multiple of 3 should be div by 3.
for a 5 digit number to be div by 3, the sum of digits (given the digits here) can be only 12 or 15.
For a sum of 12, the digits that can be used : 0,1,2,4,5
for a sum of 15: 1,2,3,4,5
Number of numbers from the first set = 4.4! (0 cannot be the first digit in the numbers)
for the second set : 5!
total = 5! +4.4! = 4!(5+4) = 24*9 = 216

Since 0 cannot be the first digit of a number, for the first position, you have 4 choices (all digits except zero). No such constraints exist for the rest of the positions; hence the next choices are 4,3,2,1 - all multiplying up to give a 4!. Had there been no 0 involved, the choices would've been 5! Instead of 4.4!

we are to form 5-digits number which is divisible by 3 from 0,1,2,3,4 and 5.
NOTE: As a rule of divisibility, the sum of digits in the number must be divisible by 3 for the number itself to be divisible by 3.
For a 5-digit numbers, only 1,2,3,4 and 5 can have their sum divisible by 3. This means that we must exclude 0 from our selection, as there is no permutation that will be divisible by 3 if 0 is among.
Therefore , Number of 5-digits number that could be formed, excluding 0 is = 5! = 5*4*3*2! = 120 possible 5-digit numbers