How many four digit numbers, which are divisible by $6$, can be formed using the digits $0, 2, 3, 4, 6$, such that no digit is used more than once and $0$ does not occur in the left-most position?

For a number to be divisible by $6$, it must satisfy both conditions:

• Divisible by $2$: Last digit must be even ($0, 2, 4,$ or $6$)

• Divisible by $3$: Sum of all digits must be divisible by $3$

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We need to pick $4$ digits from $\{0, 2, 3, 4, 6\}$. Since we're leaving out $1$ digit, let's check each possibility:

Sum of all digits $= 0 + 2 + 3 + 4 + 6 = 15$

• Leave out $0$: Sum $= 15 - 0 = 15$, and $15 \div 3 = 5$ ✓

• Leave out $2$: Sum $= 15 - 2 = 13$, and $13 \div 3 = 4.33...$ ✗

• Leave out $3$: Sum $= 15 - 3 = 12$, and $12 \div 3 = 4$ ✓

• Leave out $4$: Sum $= 15 - 4 = 11$, and $11 \div 3 = 3.66...$ ✗

• Leave out $6$: Sum $= 15 - 6 = 9$, and $9 \div 3 = 3$ ✓

Valid digit sets:

• Set A: $\{2, 3, 4, 6\}$ (leave out $0$)

• Set B: $\{0, 2, 4, 6\}$ (leave out $3$)

• Set C: $\{0, 2, 3, 4\}$ (leave out $6$)

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Set A: $\{2, 3, 4, 6\}$ (No zero restrictions!)

Last digit options: $2, 4, 6$ (all even)

For each even last digit: $3! = 6$ ways to arrange the other $3$ digits

Total: $3 \times 6 = 18$ numbers

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Set B: $\{0, 2, 4, 6\}$ (Zero cannot be first!)

Last digit can be: $0, 2, 4,$ or $6$ (all even)

When last digit is $0$:

• First digit: $3$ choices $(2, 4, 6)$

• Middle $2$ positions: $2! = 2$ arrangements

• Subtotal: $3 \times 2 = 6$

When last digit is $2$:

• First digit: $2$ choices $(4, 6)$ [since $0$ can't be first]

• Middle $2$ positions: $2! = 2$ arrangements

• Subtotal: $2 \times 2 = 4$

When last digit is $4$:

• First digit: $2$ choices $(2, 6)$ [since $0$ can't be first]

• Middle $2$ positions: $2! = 2$ arrangements

• Subtotal: $2 \times 2 = 4$

When last digit is $6$:

• First digit: $2$ choices $(2, 4)$ [since $0$ can't be first]

• Middle $2$ positions: $2! = 2$ arrangements

• Subtotal: $2 \times 2 = 4$

Set B total: $6 + 4 + 4 + 4 = 18$ numbers

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Set C: $\{0, 2, 3, 4\}$ (Zero cannot be first!)

Last digit options: $0, 2, 4$ (even digits only)

When last digit is $0$:

• First digit: $3$ choices $(2, 3, 4)$

• Middle $2$ positions: $2! = 2$ arrangements

• Subtotal: $3 \times 2 = 6$

When last digit is $2$:

• First digit: $2$ choices $(3, 4)$ [since $0$ can't be first]

• Middle $2$ positions: $2! = 2$ arrangements

• Subtotal: $2 \times 2 = 4$

When last digit is $4$:

• First digit: $2$ choices $(2, 3)$ [since $0$ can't be first]

• Middle $2$ positions: $2! = 2$ arrangements

• Subtotal: $2 \times 2 = 4$

Set C total: $6 + 4 + 4 = 14$ numbers

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Final answer: $18 + 18 + 14 = 50$