If a five digit number divisible by 3 is to be formed using the numbers $0,1,2,3,4$ and 5 without repetition…
If a five digit number divisible by 3 is to be formed using the numbers $0,1,2,3,4$ and 5 without repetition, then the total number of ways this can be done is
120
144
192
216
Solution
Any number is divisible by 3 if sum of its all digit is divisible by 3 .
(i) Sum when digits are $0,1,2,3,4=10$ not divisible by 3
(ii) Sum when digits are $0,1,2,3,5=11$ not divisible by 3
(iii) Sum when digits are $0,1,2,4,5=12$ divisible by 3
So, number of 5 digit numbers with $0,1,2,4,5$ are $4 \times 4 \times 3 \times 2 \times 1=96$
(iv) Sum when digits are $0,1,3,4,5=13$ not divisible by 3
(v) Sum when digits are $0,2,3,4,5=14$ not divisible by 3
(vi) Sum when digits are $1,2,3,4,5=15$ divisible by 3 So, number of 5 digit numbers with $1,2,3,4,5$ are $5 \times 4 \times 3 \times 2 \times 1=120$.
Thus, required no. of ways $=96+120=216$.