Example 1-16 Find the sum of integers formed by the digits $1,3,5,7$ without repetition.

 

The digits $1,3,5,7$ can form various integers where each digit appears only once in each integer. The task is to find the sum of all such possible integers.
The integer can have at most 4 digits.
1-digit numbers: 4 of them: $1,3,5,7$,
2-digit numbers: $P(4,2)=12$ of them:
$$
13,15,17,31,35,37,51,53,57,71,73,75 \text {, }
$$
3-digit numbers: $P(4,3)=24$ of them:
$$
\begin{array}{c}
135,137,315,317,513,517,713,715, \\
153,173,351,371,531,571,731,751, \\
157,175,357,375,537,573,735,753 .
\end{array}
$$
4-digit numbers: $4!=24$ of them:
$$
\begin{array}{l}
1357,1375,1537,1573,1735,1753, \\
3157,3175,3517,3571,3715,3751, \\
5137,5173,5317,5371,5713,5731, \\
7135,7153,7315,7351,7513,7531 .
\end{array}
$$

Find the sum of these numbers. Directly summing these numbers is not the goal; this is enumeration, which is only used when there is no other way. Now, let's try to find the regularity. Pay attention to the distribution of $1,3,5,7$ in the units, tens, hundreds, and thousands places. For example, in three-digit numbers, the first digit is 1 in $135,153,137,173,157,175$, where $35,53,37,73,57,75$ are permutations of two numbers from $3,5,7$.

Let $S_{1}$ represent the sum of the units, $S_{2}$ the sum of the tens, and similarly $S_{3}, S_{4}$ the sums of the hundreds and thousands places. Find $S_{1}, S_{2}, S_{3}, S_{4}$, then the sum we are looking for is
$$
S=S_{1}+10 S_{2}+100 S_{3}+1000 S_{4}
$$
(1) Calculation of $S_{1}$:

Sum of units in one-digit numbers $=1+3+5+7=16$,
Sum of units in two-digit numbers $=P(3,1) \times 16=3 \times 16=48$,
Sum of units in three-digit numbers $=P(3,2) \times 16=6 \times 16=96$,
Sum of units in four-digit numbers $=P(3,3) \times 16=6 \times 16=96$.
$$
S_{1}=16+48+96+96=256 \text {. }
$$
(2) Calculation of $S_{2}$:

Sum of tens in two-digit numbers $=P(3,1) \times 16=3 \times 16=48$.
Sum of tens in three-digit numbers $=P(3,2) \times 16=6 \times 16=96$.
Sum of tens in four-digit numbers $=P(3,3) \times 16=6 \times 16=96$.
$$
S_{2}=48+96+96=240 \text {. }
$$
(3) Calculation of $S_{3}$:

Sum of hundreds in three-digit numbers $=P(3,2) \times 16=6 \times 16=96$.
Sum of hundreds in four-digit numbers $=P(3,3) \times 16=6 \times 16=96$.
$$
S_{3}=96+96=192 \text {. }
$$
(4) Calculation of $S_{4}$:

Sum of thousands in four-digit numbers $S_{4}=P(3,3) \times 16=96$.
Therefore,
$$
\begin{aligned}
S & =256+240 \times 10+192 \times 100+96 \times 1000 \\
& =256+2400+19200+96000=117856 .
\end{aligned}
$$