Example 2.1.3 Try to find the total sum of integers formed by the digits $\{1,3,5,7\}$ without any repetition.

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Example 2.1.3 Try to find the total sum of integers formed by the digits $\{1,3,5,7\}$ without any repetition.
Analysis: Such integers can be 1 to 4 digits long, and their total number is $\mathrm{P}_{4}^{1}+\mathrm{P}_{4}^{2}+\mathrm{P}_{4}^{3}+\mathrm{P}_{4}^{4}=$ $4+12+24+24=64$. The problem is to find the sum of these 64 numbers.

Listing all 64 numbers and then finding their sum is not a good method. We will separately calculate the sum of the units digit $S_{1}$, the sum of the tens digit $S_{2}$, the sum of the hundreds digit $S_{3}$, and the sum of the thousands digit $S_{4}$ of these 64 numbers. The total sum $S=S_{1}+10 S_{2}+100 S_{3}+1000 S_{4}$.
Solution: First, calculate $S_{1}$. The sum of the units digits in one-digit numbers is $1+3+5+7=16$,
For two-digit numbers, we have $13,15,17 ; 31,35,37,51,53,57 ; 71,73,75$.
If the units digit is 1, the tens digit can be one of $3,5,7$, which gives $\mathrm{P}_{3}^{1}$ possibilities, ..., if the units digit is 7, the tens digit also has $\mathrm{P}_{3}^{1}$ possibilities,
Thus, the sum of the units digits in two-digit numbers is $(1+3+5+7) P_{3}^{1}=48$,
The sum of the units digits in three-digit numbers is $(1+3+5+7) \mathrm{P}_{3}^{2}=96$,
The sum of the units digits in four-digit numbers is $(1+3+5+7) \mathrm{P}_{3}^{3}=96$,

Therefore, $S_{1}=16+48+96+96=256$.
Next, calculate $S_{2}, S_{3}, S_{4}$.
$$
\begin{array}{l}
S_{2}=(1+3+5+7)\left(\mathrm{P}_{3}^{1}+\mathrm{P}_{3}^{2}+\mathrm{P}_{3}^{3}\right)=16 \times(3+6+6)=240, \\
S_{3}=(1+3+5+7)\left(\mathrm{P}_{3}^{2}+\mathrm{P}_{3}^{3}\right)=16 \times 12=192, \\
S_{4}=(1+3+5+7) \mathrm{P}_{3}^{3}=16 \times 6=96 .
\end{array}
$$

Thus, $S=256+2400+19200+96000=117856$.