Complex Numbers

To find the total number of 6-digit natural numbers that can be formed using the digits 1, 2, 3, and 4, where each digit must appear at least once, we can use the method of permutations and combinations.

Step 1: Determine the total number of arrangements.

First, we calculate the number of ways to arrange any of the 4 digits in 6 positions. Since repetition of digits is allowed, each position in the 6-digit number can be filled by any of the 4 digits. Thus, the total number of unrestricted arrangements is given by:

Step 2: Apply the Principle of Inclusion-Exclusion (PIE).

We need to subtract cases where at least one of the digits (1, 2, 3, or 4) is missing. Using the principle of inclusion-exclusion, we perform the following calculations:

• Single exclusion: Choose any 3 out of the 4 digits. The number of numbers that can be formed using only these 3 digits is for each choice. Since there are choices (omitting one digit each time), this gives .
• Double exclusion: Choose any 2 out of the 4 digits (effectively omitting two digits). Then, numbers can be formed using these two digits for each choice. Since there are choices, this equals .
• Triple exclusion: Pick only one digit to use all the time. This clearly only forms one number (e.g., 111111, 222222, etc.), thus numbers.

Applying PIE, we find:

Total numbers excluding at least one digit = Single exclusion - Double exclusion + Triple exclusion

Step 3: Subtract above result from the total arrangements.

The result from applying PIE tells us the number of numbers where at least one digit is missing. Therefore, the numbers where all digits appear at least once is:

Conclusion: Hence, the total number of 6-digit natural numbers that can be formed where each digit (1, 2, 3, and 4) appears at least once is 1560.