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.

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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.