Question

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

• A. $720$
• B. $240$
• C. ${}^5{C}_1{\times }^5{P}_2$
• D. $216$

Answer 1

The correct option is D $216$

we can not select 0 and 3 in the five-digit number as we have to select all the remaining 4 digits to form the number.
Therefore, we have the following cases:
Case 1) 0 is selected along with 1, 2, 4 and 5.
Since 0 cannot be placed at the leftmost i.e., 10000th position, hence we have 4 choices for the place.Therefore the 1000th place has 4 possibilities (0 and remaining 3 digits), the 100th place has 3 possibilities and so on.
Hence the total number of 5-digit numbers formed in this case are $4\times 4\times 3\times 2\times 1=96$.
Case 2) 3 is selected along with 1, 2, 4 and 5.
the five digits can be arranged in 5 places to form a number in $5!=120$ ways.
Hence the total number of ways to form a five digit number with the given condition is $96+120=216$.