1. Condition for divisibility by $6$

A number is divisible by $6$ iff it is divisible by both $2$ and $3$.

So the 5-digit number must satisfy:

• last digit is even,
• sum of digits is divisible by $3$.

2. Given digits

The available digits are $\{1,2,3,5,6,7\}$.

We must form a 5-digit number without repetition.

Since we are choosing 5 out of these 6 digits, exactly one digit is omitted.

The sum of all 6 digits is $1+2+3+5+6+7=24,$ and $24$ is divisible by $3$.

If one digit $d$ is omitted, the sum of the chosen 5 digits is $24-d.$ For divisibility by $3$, we need $24-d \equiv 0 \pmod 3.$ Since $24 \equiv 0 \pmod 3$, this requires $d \equiv 0 \pmod 3.$

Among the given digits, the digits divisible by $3$ are $3$ and $6$.

So only two cases are possible:

• omit $3$,
• omit $6$.

3. Case 1: Omit $3$

Then the digits used are $\{1,2,5,6,7\}$.

For divisibility by $2$, the last digit must be even. The even digits available are $2,6$.

So number of choices for the last digit $=2$.

After fixing the last digit, the remaining 4 digits can be arranged in $4! = 24$ ways.

Hence total numbers in this case: $2 \cdot 24 = 48.$

4. Case 2: Omit $6$

Then the digits used are $\{1,2,3,5,7\}$.

Now the only even digit available is $2$.

So the last digit must be $2$.

The remaining 4 digits can be arranged in $4! = 24$ ways.

Hence total numbers in this case: $1 \cdot 24 = 24.$

5. Total count

Therefore, total number of required 5-digit numbers is $48+24=72.$

6. Matching with options

$72$ corresponds to Option D.

7. Comparison with stored correct answer

Stored correct answer is D, and our derived answer is also D. So they agree.