Permutation and combination formation of numbers

Question

In how many ways we can make 5 digit number out of 0,1,2,3,4,7,8 such that it is divisible by 3

Answer 1

We observe the remainders given by $0,1,2,3,4,7,8$ when divided by $3$ is $0,1,-1,0,1,1,-1$.

The Rule of divisiblity by 3 is sum of digits is divisible by 3, so permutations should be such that remainders add upto $0 $ or $3$.

• Remainders adding upto 0

  • case 1: $(0,1,1,-1,-1)$ the permutations that give these set of remainders are $(0,1,4,2,8),(0,1,7,2,8),(0,7,4,2,8),(3,1,4,2,8),(3,1,7,2,8),(3,7,4,2,8)$
  • case 2: $(0,0,1,1,1)$ the permutations that give these set of remainders are $(0,3,1,4,7)$

• Hence 4 cases that contain one zero can be written as 5 digit numbers in $4\cdot 4!$ AND other 3 cases which have all non zero distinct numbers can be written as 5 digit numbers in $5!$ ways. $$\text{Multiples of 3} = 4\cdot 4 \cdot 4!+3\cdot 5!=744$$