To solve this problem, we need to count the number of 4-digit numbers greater than 3600 that can be formed using the digits $0, 1, 3, 5, 7, 9$ without repetition. Let us proceed step by step:

Step 1: Criteria for a number to be greater than 3600

A 4-digit number is greater than 3600 if:

1. Its thousands place digit is at least 3 (i.e., $3, 5, 7, 9$).
2. No digit is repeated.

Step 2: Counting possibilities

Case 1: Thousands place = 3

• Fix the thousands place as $3$.
• Remaining digits: $\{ 0, 1, 5, 7, 9 \}$.
• Hundreds place: Can be any of the 5 remaining digits $\{ 0, 1, 5, 7, 9 \}$ → 5 choices.
• Tens place: Can be any of the 4 remaining digits → 4 choices.
• Units place: Can be any of the 3 remaining digits → 3 choices.

The total numbers for this case = $1 \cdot 5 \cdot 4 \cdot 3 = 60$.

Case 2: Thousands place = 5

• Fix the thousands place as $5$.
• Remaining digits: $\{ 0, 1, 3, 7, 9 \}$.
• Hundreds place: Can be any of the 5 remaining digits $\{ 0, 1, 3, 7, 9 \}$ → 5 choices.
• Tens place: Can be any of the 4 remaining digits → 4 choices.
• Units place: Can be any of the 3 remaining digits → 3 choices.

The total numbers for this case = $1 \cdot 5 \cdot 4 \cdot 3 = 60$.

Case 3: Thousands place = 7

• Fix the thousands place as $7$.
• Remaining digits: $\{ 0, 1, 3, 5, 9 \}$.
• Hundreds place: Can be any of the 5 remaining digits $\{ 0, 1, 3, 5, 9 \}$ → 5 choices.
• Tens place: Can be any of the 4 remaining digits → 4 choices.
• Units place: Can be any of the 3 remaining digits → 3 choices.

The total numbers for this case = $1 \cdot 5 \cdot 4 \cdot 3 = 60$.

Case 4: Thousands place = 9

• Fix the thousands place as $9$.
• Remaining digits: $\{ 0, 1, 3, 5, 7 \}$.
• Hundreds place: Can be any of the 5 remaining digits $\{ 0, 1, 3, 5, 7 \}$ → 5 choices.
• Tens place: Can be any of the 4 remaining digits → 4 choices.
• Units place: Can be any of the 3 remaining digits → 3 choices.

The total numbers for this case = $1 \cdot 5 \cdot 4 \cdot 3 = 60$.

Step 3: Total numbers greater than 3600

Adding up all the cases: $60 + 60 + 60 + 60 = 240$

Final Answer:

The total number of 4-digit numbers greater than 3600 that can be formed is 240.

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Would you like a detailed explanation of any step or related queries? 😊
Here are 5 related questions for practice:

1. How many 4-digit numbers greater than 5000 can be formed using $0, 1, 3, 5, 7, 9$?
2. How many 3-digit numbers can be formed using $0, 1, 3, 5, 7, 9$ without repetition?
3. If repetition is allowed, how many 4-digit numbers greater than 3600 can be formed?
4. How many even 4-digit numbers can be formed using the digits $0, 1, 3, 5, 7, 9$ without repetition?
5. What is the total number of distinct 4-digit numbers that can be formed using these digits?

Tip: Always consider restrictions on the digits (e.g., leading zeros) when solving permutation problems!