Key Concepts and Formulas

• Casework: Breaking down a problem into mutually exclusive and exhaustive cases to simplify counting.
• Fundamental Principle of Counting (Multiplication Principle): If there are $n_1$ ways to do one thing, $n_2$ ways to do another, and so on, then there are $n_1 \times n_2 \times \dots$ ways to do all of them.

Step-by-Step Solution

Step 1: Define the problem and the set of digits.

We need to find the number of four-digit numbers strictly greater than 4321 using the digits 0, 1, 2, 3, 4, and 5, with repetition allowed. Let the four-digit number be represented as $d_1 d_2 d_3 d_4$. The set of allowed digits is $S = \{0, 1, 2, 3, 4, 5\}$.

Step 2: Case 1: $d_1 > 4$

If the first digit ($d_1$) is greater than 4, the number will always be greater than 4321.

• $d_1$ can only be 5 (1 choice).
• $d_2, d_3,$ and $d_4$ can be any digit from the set $S$ (6 choices each).
• Number of possibilities in this case: $1 \times 6 \times 6 \times 6 = 216$.

Step 3: Case 2: $d_1 = 4$ and $d_2 > 3$

If $d_1$ is 4, then for the number to be greater than 4321, $d_2$ must be greater than 3.

• $d_1$ must be 4 (1 choice).
• $d_2$ can be 4 or 5 (2 choices).
• $d_3$ and $d_4$ can be any digit from the set $S$ (6 choices each).
• Number of possibilities in this case: $1 \times 2 \times 6 \times 6 = 72$.

Step 4: Case 3: $d_1 = 4$, $d_2 = 3$ and $d_3 > 2$

If $d_1$ is 4 and $d_2$ is 3, then for the number to be greater than 4321, $d_3$ must be greater than 2.

• $d_1$ must be 4 (1 choice).
• $d_2$ must be 3 (1 choice).
• $d_3$ can be 3, 4, or 5 (3 choices).
• $d_4$ can be any digit from the set $S$ (6 choices).
• Number of possibilities in this case: $1 \times 1 \times 3 \times 6 = 18$.

Step 5: Case 4: $d_1 = 4$, $d_2 = 3$, $d_3 = 2$ and $d_4 > 1$

If $d_1$ is 4, $d_2$ is 3, and $d_3$ is 2, then for the number to be greater than 4321, $d_4$ must be greater than 1.

• $d_1$ must be 4 (1 choice).
• $d_2$ must be 3 (1 choice).
• $d_3$ must be 2 (1 choice).
• $d_4$ can be 2, 3, 4, or 5 (4 choices).
• Number of possibilities in this case: $1 \times 1 \times 1 \times 4 = 4$.

Step 6: Calculate the total number of possibilities.

Add the number of possibilities from each case: Total = $216 + 72 + 18 + 4 = 310$

Common Mistakes & Tips

• Overlapping Cases: Ensure that the cases are mutually exclusive. Avoid double-counting any numbers.
• Missing Cases: Make sure that all possible scenarios are covered to avoid undercounting.
• Zero as the first digit: Be mindful of the restriction that the first digit of a four-digit number cannot be zero, although it is not applicable in this problem since we are looking for numbers greater than 4321.

Summary

We used casework to break down the problem into manageable scenarios based on the digits of the four-digit number. By considering each digit from left to right and ensuring the number is strictly greater than 4321, we found the total number of such numbers by summing the possibilities from each case. The total number of four-digit numbers strictly greater than 4321 that can be formed using the digits 0, 1, 2, 3, 4, 5 with repetition allowed is 310.

The final answer is $\boxed{310}$, which corresponds to option (C).