A five-digit number divisible by $3$ is to be formed using the numbers $0,1,2,3,4,5$ without repetitions. The total number of ways this can be done is:

Sum of digits $0+1+2+3+4+5 = 15$, which is divisible by $3$. If we remove one digit, the remaining sum must also be divisible by $3$ for divisibility. Possible removals giving divisible sums: - Remove $0$ → sum $15$ ✔ - Remove $3$ → sum $12$ ✔ - Remove $6$ → not present - Remove others → not divisible. Hence, we can form numbers using digits $\{1,2,3,4,5\}$ and $\{0,1,2,4,5\}$. Case 1: Using $\{1,2,3,4,5\}$ — 5 digits, all used, so $5! = 120$. Case 2: Using $\{0,1,2,4,5\}$ — first digit cannot be 0, so $4\times4! = 96$. Total = $120 + 96 = 216$.