Correct Answer: Option B ($1440$)
Step 1: Identify Constraints
The numbers must be greater than $2000$ using the set $S = \{0, 1, 2, 3, 4, 5\}$. Since the digits cannot be repeated, we must consider all possible valid "lengths" of the integers. We can form $4$-digit, $5$-digit, and $6$-digit numbers. Any $7$-digit number would require repetition, which is prohibited.
Step 2: Calculate Cases
Case 1: $4$-digit numbers
To be greater than $2000$, the thousands place must be $2, 3, 4,$ or $5$.Thousands place: $4$ options $\{2, 3, 4, 5\}$.Hundreds place: $5$ remaining options (including $0$ and $1$).Tens place: $4$ remaining options.Units place: $3$ remaining options.Total $4$-digit integers $= 4 \times 5 \times 4 \times 3 = 240$.
Case 2: $5$-digit numbers
Any $5$-digit number formed using these digits will inherently be greater than $2000$.
First digit: $5$ options (cannot be $0$).
Second digit: $5$ options (remaining digits plus $0$).
Third digit: $4$ options.
Fourth digit: $3$ options.
Fifth digit: $2$ options.
Total $5$-digit integers $= 5 \times 5 \times 4 \times 3 \times 2 = 600$.
Case 3: $6$-digit numbers
Any $6$-digit number formed using these digits will also be greater than $2000$.
First digit: $5$ options (cannot be $0$).
Remaining $5$ positions can be filled in $5!$ ways.
Total $6$-digit integers $= 5 \times 120 = 600$.
Step 3: Final Computation
Summing the mutually exclusive cases:$$\text{Total} = 240 + 600 + 600 = 1440$$