Key Concepts and Formulas

• Permutations: The number of ways to arrange $r$ objects from a set of $n$ distinct objects, where order matters, is given by $P(n, r) = \frac{n!}{(n-r)!}$.
• Divisibility Rule for 5: A number is divisible by 5 if and only if its last digit is 0 or 5.
• Casework: When faced with multiple constraints that interact, consider breaking the problem into distinct cases to handle each constraint systematically.

Step-by-Step Solution

Step 1: Analyze the constraints and identify the most restrictive positions.

We are forming a five-digit number $d_1d_2d_3d_4d_5$ using the digits $\{0, 1, 3, 5, 7, 9\}$ without repetition. The constraints are: the number must be greater than 40000, and divisible by 5. The first digit $d_1$ and the last digit $d_5$ are the most constrained. We will examine the possible choices for these digits.

Step 2: Determine the possible choices for the first digit, $d_1$.

Since the number must be greater than 40000, $d_1$ must be greater than or equal to 4. Also, $d_1$ cannot be 0 (as it's a five-digit number). From the given digits, the possible values for $d_1$ are 5, 7, and 9. Thus, $d_1 \in \{5, 7, 9\}$. There are 3 possible choices for $d_1$.

Step 3: Determine the possible choices for the last digit, $d_5$.

Since the number must be divisible by 5, the last digit $d_5$ must be either 0 or 5. Thus, $d_5 \in \{0, 5\}$. There are 2 possible choices for $d_5$.

Step 4: Consider the cases based on the value of $d_5$ to account for the "without repetition" constraint.

We need to consider two cases based on the value of the last digit $d_5$:

• Case A: $d_5 = 0$. If $d_5 = 0$, then $d_1$ can be any of the digits in $\{5, 7, 9\}$. So, there are 3 choices for $d_1$. The possible pairs for $(d_1, d_5)$ are $(5, 0), (7, 0), (9, 0)$.

• Case B: $d_5 = 5$. If $d_5 = 5$, then $d_1$ can be either 7 or 9 (since it cannot be 5 because of the "without repetition" constraint). So, there are 2 choices for $d_1$. The possible pairs for $(d_1, d_5)$ are $(7, 5), (9, 5)$.

Step 5: Calculate the total number of ways to choose $d_1$ and $d_5$.

The total number of ways to choose the first and last digits is the sum of the possibilities from both cases: $3 + 2 = 5$.

Step 6: Determine the number of ways to fill the remaining three digits, $d_2, d_3,$ and $d_4$.

After choosing $d_1$ and $d_5$, we have 4 digits remaining from the original set of 6 digits. We need to arrange these 4 digits into the 3 remaining positions ($d_2, d_3, d_4$). This is a permutation problem since the order matters. The number of ways to arrange 3 digits out of 4 is $P(4, 3) = \frac{4!}{(4-3)!} = \frac{4!}{1!} = 4 \times 3 \times 2 = 24$.

Step 7: Calculate the total number of five-digit numbers satisfying the given conditions.

To find the total number of such numbers, we multiply the number of ways to choose the first and last digits by the number of ways to arrange the middle three digits: $5 \times 24 = 120$.

Common Mistakes & Tips

• Forgetting the "without repetition" constraint: This is the most common mistake. Always consider the impact of choosing a digit on the subsequent choices.
• Incorrectly calculating permutations: Make sure to use the correct permutation formula and understand when order matters.
• Not using casework: When constraints interact, casework can help to systematically address all possibilities.

Summary

We analyzed the constraints on the first and last digits of the five-digit number, considering the "without repetition" rule through casework. This resulted in 5 possible ways to choose the first and last digits. For each of these 5 ways, the remaining 3 middle positions could be filled in $P(4,3) = 24$ ways. Multiplying these possibilities, we obtained a total of $5 \times 24 = 120$ five-digit numbers.

Final Answer

The final answer is \boxed{120}, which corresponds to option (C).