Understanding the Problem: Finding the Sum of 4-Digit Numbers

The question asks us to find the sum of all possible four-digit numbers that can be formed using the digits 0, 1, 4, 5, with the condition that each digit must be used exactly once (no repetition) and the number must be a four-digit number (meaning the first digit cannot be 0).

Key Concepts for Calculation

To solve this, we'll use the principles of permutations and place value.

• Permutations: We are arranging the given digits. The total number of ways to arrange \(n\) distinct items is \(n!\).
• Place Value: The value of a digit depends on its position (thousands, hundreds, tens, ones).
• Constraint: A four-digit number cannot start with 0.

Step-by-Step Calculation

1. Total Possible Arrangements (Permutations)

We have 4 distinct digits: {0, 1, 4, 5}.

The total number of ways to arrange these 4 digits is \(4!\) (4 factorial):

\(4! = 4 \times 3 \times 2 \times 1 = 24\)

This gives us 24 possible arrangements.

2. Arrangements that form Valid 4-Digit Numbers

A four-digit number cannot begin with 0. We need to find how many of the 24 arrangements start with 0.

If 0 is fixed in the first position (thousands place), we need to arrange the remaining 3 digits {1, 4, 5}. The number of ways to do this is \(3!\):

\(3! = 3 \times 2 \times 1 = 6\)

So, there are 6 arrangements that start with 0 (e.g., 0145, 0154, etc.). These are not valid four-digit numbers.

The number of valid four-digit numbers is the total arrangements minus the arrangements starting with 0:

\(\text{Number of valid 4-digit numbers} = 24 - 6 = 18\)

3. Calculating the Sum of All Valid 4-Digit Numbers

We can calculate the sum by considering the contribution of each digit in each place value across all 18 valid numbers.

Method: Sum of all permutations minus sum of permutations starting with 0.

a) Sum of all 24 permutations (treating them as numbers):

Let the sum of the digits be \(S = 0 + 1 + 4 + 5 = 10\).

In the 24 permutations, each digit (0, 1, 4, 5) appears in each place value (thousands, hundreds, tens, ones) an equal number of times. The number of times each digit appears in a specific position is \((4-1)! = 3! = 6\) times.

The sum of all permutations is calculated as:

\(\text{Sum} = S \times (n-1)! \times (1111)\)

Where \(n=4\) (number of digits).

\(\text{Sum}_{\text{all}} = 10 \times 6 \times 1111 = 60 \times 1111 = 66660\)

This sum (66660) includes the values of arrangements starting with 0 (like 0145 treated as 145).

b) Sum of the 6 arrangements starting with 0:

These are permutations where 0 is fixed in the thousands place. The remaining digits {1, 4, 5} form the hundreds, tens, and ones places. The numbers formed are essentially 3-digit numbers using {1, 4, 5}.

The sum of these digits is \(S' = 1 + 4 + 5 = 10\).

There are \(3! = 6\) such numbers.

Each digit (1, 4, 5) appears in the hundreds, tens, and ones places \((3-1)! = 2! = 2\) times.

The sum of these 3-digit numbers is:

\(\text{Sum}_{\text{0-start}} = S' \times (n-1)! \times (111)\)

Where \(n=3\) (for the digits {1, 4, 5}).

\(\text{Sum}_{\text{0-start}} = 10 \times 2 \times 111 = 20 \times 111 = 2220\)

c) Sum of valid 4-Digit Numbers:

To get the sum of only the valid four-digit numbers, we subtract the sum of the numbers starting with 0 from the total sum of all permutations.

\(\text{Sum}_{\text{valid}} = \text{Sum}_{\text{all}} - \text{Sum}_{\text{0-start}}\)

\(\text{Sum}_{\text{valid}} = 66660 - 2220 = 64440\)

Conclusion

The sum of all four-digit numbers formed using the digits 0, 1, 4, 5 without repetition is 64440.