How many five-digit numbers can be formed with the digits 5, 4, 3, 5, 3?

Explanation

To solve the problem of how many five-digit numbers can be formed with the digits 5, 4, 3, 5, 3, we will follow these steps:

### Step 1: Identify the digits and their frequencies
The digits we have are: 5, 4, 3, 5, 3.
- The digit 5 occurs 2 times.
- The digit 4 occurs 1 time.
- The digit 3 occurs 2 times.

### Step 2: Use the formula for permutations of multiset
The formula for the number of distinct permutations of a multiset is given by:

\[ \text{Number of permutations} = \frac{n!}{n_1! \cdot n_2! \cdot n_3! \cdots} \]

where:
- \( n \) is the total number of items,
- \( n_1, n_2, n_3, \ldots \) are the frequencies of the distinct items.

In our case:
- Total digits \( n = 5 \) (since we have five digits: 5, 4, 3, 5, 3)
- Frequencies: \( n_1 = 2 \) (for digit 5), \( n_2 = 1 \) (for digit 4), \( n_3 = 2 \) (for digit 3)

### Step 3: Substitute into the formula
Substituting the values into the formula, we get:

\[ \text{Number of permutations} = \frac{5!}{2! \cdot 1! \cdot 2!} \]

### Step 4: Calculate the factorials
Now, we calculate the factorials:
- \( 5! = 120 \)
- \( 2! = 2 \)
- \( 1! = 1 \)

### Step 5: Substitute the factorial values
Now substituting these values back into the formula:

\[ \text{Number of permutations} = \frac{120}{2 \cdot 1 \cdot 2} = \frac{120}{4} = 30 \]

### Step 6: Conclusion
Thus, the total number of distinct five-digit numbers that can be formed with the digits 5, 4, 3, 5, 3 is **30**.

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