Question

A five digit number divisible by $3$ is to be formed using the numerals $0,1,2,3,4$ and $5$ without repetition. The total number of ways in which this can be done is:

• A. $211$
• B. $216$
• C. $221$
• D. $311$

Answer 1

The correct option is B $216$

All permutations formed with $1,2,3,4,5$ (sum $=15$) will be divisible by $3$.
There are $5!=120$ such permutations.

Such numbers can also be formed using $0$ and $1,2,3,4,5$.

There are $4\times 4!$ such numbers, i.e. $96$.

(Factors of $4$ for four positions of $0$ and $4!$ for different permutations of these four numbers)
$\therefore$ Total of such numbers $=120+96=216$