Answer: Option A
:
A
We know that a five digit number is divisible by 3, if and only if sum of its digits (= 15) is divisible by 3, therefore we should not use 0 or 3 while forming the five digit numbers.
Now,
(i) In case we do not use 0 the five digit number can be formed (from the digit 1, 2, 3, 4, 5) in ${}^5{P}_5$ways.
(ii) In case we do not use 3, the five digit number can be formed (from the digit 0, 1, 2, 4, 5) in ${}^5{P}_5-{}^4{P}_4$= 5! - 4!= 120 - 24 = 96 ways.
(${}^4{P}_4=$ cases when $0$ is at the first position)
$\therefore$The total number of such 5 digit number = ${}^5{P}_5+({}^5{P}_5-{}^4{P}_4)=120+96=216$