Concept:A number divisible by 6 must be divisible by 2 and 3.
Since the sum of digits
0+1+2+3+4+5=15 is always divisible by 3, any number formed using all six digits is divisible by 3.
Only the last digit (units place) must be even for divisibility by 2.
Explanation:We form 6-digit numbers using each digit exactly once.
The first digit cannot be 0 (otherwise it is a 5-digit number).
Last digit must be even: 0, 2, or 4 (three cases).
Case 1: Last digit = 0First digit: any of the remaining 5 digits (1,2,3,4,5) — 5 choices.
Remaining 4 positions: arrange the other 4 digits in
4! ways.
Number of such numbers =
5×4!=5×24=120.
Case 2: Last digit = 2First digit: cannot be 0 or 2, so choices from {1,3,4,5} — 4 choices.
Then the remaining 4 positions: fill with the leftover 4 digits (including 0) in
4!=24 ways.
Number =
4×24=96.
Case 3: Last digit = 4First digit: cannot be 0 or 4, so choices from {1,2,3,5} — 4 choices.
Remaining 4 positions:
4!=24 ways.
Number =
4×24=96.
Total valid numbers =
120+96+96=312.
Answer:312 (Option D)