A five digit number divisible by 3 is to be formed using the digits $0,1,2,3,4,5$ without repetition, then…
A five digit number divisible by 3 is to be formed using the digits $0,1,2,3,4,5$ without repetition, then the total number of ways this can be done is
216
240
96
120
Solution
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 and 3 in a same number while forming the five digit numbers.
Now,
i. In case we do not use 0 , the five digit number can be formed (using digits $1,2,3,4,5$ ) in ${ }^5 \mathrm{P}_5=120$ ways.
ii. In case we do not use 3, the five digit number can be formed (using digits $0,1,2,4,5$ ) in ${ }^5 \mathrm{P}_5-{ }^4 \mathrm{P}_4=5!-4!=120-24=96$ ways
$\therefore \quad$ The total number of such 5 digit number $=120+96=216$