To form a 6-digit number using the digits 4, 5, 8, and 9, where two of these digits occur twice and the other two occur once, we can approach the problem step by step.

Step 1: Choose the Digits that Occur Twice

• There are 4 digits (4, 5, 8, 9) available.
• We need to choose 2 out of these 4 digits to repeat twice.
• The number of ways to choose 2 digits from 4 is calculated using the combination formula $\binom{4}{2}$.

$\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 \text{ ways}$

Step 2: Arrange the Digits

• Once we have selected the two digits that will repeat, we need to arrange the six digits.
• The digits are: $a, a, b, b, c, d$ where $a$ and $b$ are the digits that repeat, and $c$ and $d$ are the remaining digits.
• The number of ways to arrange 6 items where 2 items are identical, another 2 are identical, and the last 2 are distinct is given by:

$\frac{6!}{2! \times 2!} = \frac{720}{4} = 180 \text{ ways}$

Step 3: Calculate the Total Number of 6-Digit Numbers

• For each choice of 2 digits to repeat, there are 180 possible arrangements.
• Since there are 6 ways to choose which digits will repeat, the total number of different 6-digit numbers is:

$6 \times 180 = 1080 \text{ different 6-digit numbers}$

So, there are 1,080 different ways to form a 6-digit number using the digits 4, 5, 8, and 9 where two of the digits occur twice.

---

Would you like more details on this calculation or have any questions?

Related Questions:

1. How would the total change if all four digits could repeat any number of times?
2. How many 6-digit numbers can be formed if no digit is allowed to repeat?
3. What if the number formed must be divisible by a specific digit, like 5?
4. Can you derive the number of possible 7-digit numbers with similar conditions?
5. What is the probability of randomly selecting a number that starts with 4?

Tip:

When working with combinations and permutations, clearly distinguish between cases where order matters and where it doesn't.