You have the digits 2, 2, 3, 3, 4, 5. How many unique four-digit numbers can you form using these digits?

can we solve it as pairs and distincts instead of having many cases

Yes, we can solve this problem by categorizing the digits into "pairs" and "distinct" digits to simplify the counting process. Let's break down the problem accordingly:

### Given Digits:
- Two pairs: \(2, 2\) and \(3, 3\)
- Two distinct digits: \(4\) and \(5\)

### Forming a Four-Digit Number:
We have the following scenarios:

1. **Two digits from pairs and two distinct digits:**
   - Choose 1 digit from each pair: \(2\) and \(3\)
   - Use the distinct digits: \(4\) and \(5\)
   - Number of arrangements: \( \frac{4!}{1!1!1!1!} = 24 \)

2. **One digit from a pair and three distinct digits:**
   - Choose one pair (either \(2, 2\) or \(3, 3\)) and use one digit from it, then choose three distinct digits from \(2, 3, 4, 5\).
   
   Since we have only two pairs (2, 2 and 3, 3), we consider them separately:

   **Case 1: Choose 2 and distinct digits 3, 4, 5**
   - Number of arrangements: \( \frac{4!}{2!} = 12 \)
   
   **Case 2: Choose 3 and distinct digits 2, 4, 5**
   - Number of arrangements: \( \frac{4!}{2!} = 12 \)
   
   Total for this scenario: \(12 + 12 = 24\)

3. **Two identical digits and two distinct digits (e.g., \(22xx\)):**
   - Choose 2 identical digits (either \(2, 2\) or \(3, 3\)) and 2 distinct digits from the remaining four.
   
   For each pair (2, 2) and (3, 3):
   
   **Case 1: \(2, 2\) with distinct digits from {3, 4, 5}**
   - Number of arrangements: \( \frac{4!}{2!1!1!} = 12 \)
   
   **Case 2: \(3, 3\) with distinct digits from {2, 4, 5}**
   - Number of arrangements: \( \frac{4!}{2!1!1!} = 12 \)
   
   Total for this scenario: \(12 + 12 = 24\)

### Final Calculation:
Summing up all scenarios:

\[
24 \,(\text{two digits from pairs and two distinct}) + 24 \,(\text{one digit from pair and three distinct}) + 24 \,(\text{two identical digits and two distinct}) = 72
\]

Therefore, the number of unique four-digit numbers that can be formed using these digits is **72**.

Would you like to see further details or have any questions?

Here are some follow-up questions:

1. What happens if we add another distinct digit?
2. How does the solution change if we use only one pair?
3. What if we are required to form a five-digit number instead?
4. How would the solution differ if we have three pairs?
5. How can we generalize this method for any set of digits?
6. Can we approach the problem using a combinatorial identity?
7. How does the problem change with repetition of digits allowed?
8. Can we use the digits to form any palindromes?

**Tip:** When solving combinatorial problems, organizing cases systematically helps to avoid missing any possibilities or overcounting.

Explore how to count unique four-digit numbers using the digits 2, 2, 3, 3, 4, 5. This mathematical problem involves combinatorics and permutations to determine various combinations of pairs and distinct digits. Learn the detailed step-by-step solution with clear explanations and calculations.

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