We need to count integers greater than $6000$ formed using the digits $3,5,6,7,8$ without repetition.

Since the number must be greater than $6000$, it can have either:

• $4$ digits, or
• $5$ digits.

Count 4-digit numbers greater than $6000$

For a 4-digit number to be greater than $6000$, its thousand's digit must be one of $6,7,8$.

• Choose the first digit: $3$ ways ($6,7,8$)
• Choose and arrange the remaining $3$ places from the remaining $4$ digits: $^4P_3 = 4 \times 3 \times 2 = 24$

So, number of such 4-digit numbers is $3 \times 24 = 72$

Count 5-digit numbers

Any 5-digit number formed from these digits is automatically greater than $6000$.

Number of 5-digit numbers using all $5$ digits without repetition: $5! = 120$

Total numbers $72 + 120 = 192$

Therefore, the required number of integers is $\boxed{192}$

Comparison with stored answer

Stored correct answer is D, which corresponds to $192$.

Hence, my derived answer agrees with the stored answer.