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The question asks: **How many odd numbers, greater than 500,000, can be made from the digits 2, 3, 4, 5, 6, 7 without repetitions?**

Let's break it down step by step:

### Step 1: Identify the constraints
1. **Odd number**: For a number to be odd, its last digit must be odd. From the available digits (2, 3, 4, 5, 6, 7), the odd digits are **3, 5, and 7**. So, the last digit must be one of these three.
   
2. **Greater than 500,000**: The number must be a 6-digit number, and its first digit must be 5, 6, or 7 to ensure it is greater than 500,000.

### Step 2: Number structure and choices
We need to construct a 6-digit odd number that satisfies the constraints. We'll handle it by considering the available choices for each digit step-by-step:

- **First digit**: The first digit must be **5, 6, or 7** to satisfy the condition that the number is greater than 500,000. Therefore, there are **3 choices** for the first digit.
  
- **Last digit**: The last digit must be odd, so it can be **3, 5, or 7**. There are **3 choices** for the last digit.

- **Remaining four digits**: After choosing the first and last digits, we have four remaining digits to fill the middle four positions. Since repetitions are not allowed, and two digits have already been used, we will choose from the remaining four digits for each middle position:

  - For the second position, there are **4 remaining choices**.
  - For the third position, there are **3 remaining choices**.
  - For the fourth position, there are **2 remaining choices**.
  - For the fifth position, there is **1 remaining choice**.

### Step 3: Total number of combinations
Now, we multiply the number of choices for each position:

- First digit: \(3\) choices (5, 6, or 7)
- Second digit: \(4\) choices
- Third digit: \(3\) choices
- Fourth digit: \(2\) choices
- Fifth digit: \(1\) choice
- Last digit: \(3\) choices (3, 5, or 7)

The total number of odd numbers greater than 500,000 is:
\[
3 \times 4 \times 3 \times 2 \times 1 \times 3 = 216
\]

### Step 4: Exclusion of invalid cases
We have a possible issue if the first and last digit turn out to be the same number. Specifically:
- If the first digit is 5, the last digit cannot also be 5.
- If the first digit is 7, the last digit cannot also be 7.

Let's count these invalid cases:
- **First digit is 5 and last digit is 5**: The remaining four digits can still be filled with the 4 other available digits, leading to \( 4 \times 3 \times 2 \times 1 = 24 \) numbers.
- **First digit is 7 and last digit is 7**: This again leads to \( 24 \) numbers.

Thus, we need to subtract these 48 invalid cases from the total:

\[
216 - 48 = 168
\]

### Final Answer:
The total number of odd numbers greater than 500,000 that can be made from the digits 2, 3, 4, 5, 6, 7 without repetition is **168**.

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Would you like further details or have any questions?

Here are 5 related questions:
1. How would the answer change if the condition for "greater than 500,000" was not included?
2. How many even numbers greater than 500,000 can be made under the same conditions?
3. What if the digits could be repeated? How many odd numbers greater than 500,000 could be formed then?
4. What if the number had to be less than 500,000? How would the solution differ?
5. How many 5-digit odd numbers can be made from the digits 2, 3, 4, 5, 6, 7 without repetition?

**Tip**: When solving counting problems, carefully identifying constraints and conditions for valid cases is essential for accurate results.

How many odd numbers greater than 500,000 can be made from the digits 2, 3, 4, 5, 6, 7 without repetitions?

Learn how to calculate how many odd numbers greater than 500,000 can be formed from the digits 2, 3, 4, 5, 6, and 7 without repetitions. This combinatorics problem focuses on permutations and odd number constraints.

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