A five digit number divisible by 6 is to be formed by using the digits 0, 1, 2, 3, 4 and 8 without repetition. The total number of ways in which this can be done is

Since $0 + 1 + 2 + 3 + 4 + 8 = 18,$ the $5$ digit number will be divisible by 3 if either 0 or 3 is not used. When 0 is not used. For the unit place we have 3 choices$(2, 4 or 8)$ and for the remaining place we have $4!$ Choices.

When 3 is not used  In this case, 0 is used at the unit’s place, the number of choices is $4!$ If 0 is not used at the unit’s place, then unit’s place can be filled up in 3 ways and the remaining places in $(4! – 3!)$ ways. Thus, required number of numbers

$=3(4!)+4!+3(4!-3!)=150$