【Example 5】Find the sum of all four-digit numbers composed of $1,2,3,4,5$ without any repeated digits.



Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Solving: For this problem, it is not enough to just find out how many such four-digit numbers there are. Of course, it is also not feasible to list them all out one by one and add them up; we must think of other methods. Now, let's consider how many of these four-digit numbers start with 1? This is not a difficult problem to solve, as it is equivalent to finding out how many different ways there are to select 3 numbers from 2, 3, 4, 5 to serve as the hundreds, tens, and units digits, which is a permutation problem. Therefore, there are \( A_{4}^{3} = 4 \times 3 \times 2 = 24 \) such numbers. Applying the same reasoning to 2, 3, 4, 5, we know that the number of such four-digit numbers starting with each of them is also 24. Thus, the sum of the thousands digits of all these four-digit numbers is
\[
24 \times (1 + 2 + 3 + 4 + 5) \times 1000 = 360000
\]

Similarly, the four-digit numbers with 1, 2, 3, 4, 5 as the hundreds, tens, and units digits each have 24 such numbers. Therefore, the sum of all these four-digit numbers is
\[
24 \times (1 + 2 + 3 + 4 + 5) \times (10^3 + 10^2 + 10 + 1) = 399960
\]

Thus, this more complex summation problem has been solved by studying the patterns of the numbers and utilizing permutation methods.