The problem asks how many four-digit natural numbers can be formed using only the digits 1, 2, 3, and 4, if each of these digits can only be used once.

Step-by-step solution:

1. First digit: Since it is a four-digit number, the first digit can be any of 1, 2, 3, or 4. So, there are 4 possible choices for the first digit.

2. Second digit: After choosing the first digit, we are left with 3 digits. So, there are 3 possible choices for the second digit.

3. Third digit: Once the first two digits are chosen, 2 digits remain. Therefore, there are 2 choices for the third digit.

4. Fourth digit: After choosing the first three digits, only 1 digit is left, so there is 1 choice for the fourth digit.

Calculation:

$4 \times 3 \times 2 \times 1 = 24$

So, 24 four-digit natural numbers can be formed using the digits 1, 2, 3, and 4 if each digit is used only once.

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

Here are 5 related questions:

1. What if repetition of digits was allowed? How would the answer change?
2. How many three-digit numbers can be formed under the same condition?
3. What if we could use digits 0, 1, 2, 3, and 4 but still no repetition?
4. What if we could form five-digit numbers with digits 1, 2, 3, 4, and 5 without repetition?
5. What is the probability that a randomly chosen four-digit number formed from 1, 2, 3, and 4 is even?

Tip: Problems involving permutations like this one often require a good understanding of factorials!