can i do it like this choose two places from six places by 6C2 fill it by same number number in 4 ways then choose from remaining 4 places 2 places in 4C2 ways and fill it by same digits in 3 ways then remaining 2 places choose by 2C2 and fill it by 2 digits in 2! ways so total arrangement are 6C2*4*4C2*3*2C2*2!


Your approach is on the right track, but there are some corrections needed in the reasoning and steps. Let's go through your logic and refine it.

### Step-by-Step Explanation:

1. **Choose Two Positions for the First Repeating Digit:**
   - You correctly identified that you need to choose 2 places out of 6 for the first repeating digit.
   - The number of ways to choose 2 positions from 6 is given by \( \binom{6}{2} \).

   \[
   \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15
   \]

2. **Fill These Positions with the First Digit:**
   - You have 4 possible digits (4, 5, 8, 9) to fill these two positions.
   - So, there are 4 ways to choose this digit.

   \[
   4 \text{ ways}
   \]

3. **Choose Two Positions for the Second Repeating Digit:**
   - Now, you have 4 remaining positions, and you need to choose 2 positions for the second repeating digit.
   - The number of ways to choose 2 positions from 4 is given by \( \binom{4}{2} \).

   \[
   \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6
   \]

4. **Fill These Positions with the Second Digit:**
   - After choosing the positions, you have 3 remaining digits (excluding the digit used in Step 2) to fill these two positions.
   - So, there are 3 ways to choose this digit.

   \[
   3 \text{ ways}
   \]

5. **Choose the Last Two Positions and Fill Them with the Remaining Two Digits:**
   - The last 2 positions are automatically filled by the remaining two digits.
   - The number of ways to arrange these two digits in the last two positions is \( 2! \).

   \[
   2! = 2
   \]

### Putting It All Together:

Now, let's multiply all these steps together:

\[
\text{Total arrangements} = \binom{6}{2} \times 4 \times \binom{4}{2} \times 3 \times 2! = 15 \times 4 \times 6 \times 3 \times 2 = 2160
\]

**Final Answer:** There are 2160 possible 6-digit numbers that can be formed using the digits 4, 5, 8, and 9, where two of the digits occur twice.

### Clarification:
Your initial approach was mostly correct, but there was a misunderstanding in the final calculation. The correct total is **2160**, not 1080.

---

Would you like further clarification on any part of this process?

### Related Questions:
1. How would the result change if one digit could appear three times?
2. How many 7-digit numbers can be formed under similar conditions?
3. What if the number formed must start with a specific digit?
4. Can you calculate how many of these numbers are even?
5. What if no digit is allowed to repeat more than once?

### Tip:
Double-check each step in combinatorial problems to ensure all factors are accounted for and correctly applied.

Understand how to calculate the total number of 6-digit combinations using repeating digits. This mathematical problem involves combinatorics, permutations, and factorial calculations. Learn step-by-step how to determine the total arrangements of digits 4, 5, 8, and 9 with specific repetitions.

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