Let $S$ be the set numbers whose digits are chosen from ${1, 3, 5, 7}$ such that no digits are repeated. Find the sum of every element in $S$. All numbers in $S$ are natural. I could find the $|S| = 64$ by my own. Can't find the sum of every number in $S$, nor understand the book's explanation for that. The answer is $117856$.
Taken from Principles and Techniques in Combinatorics.
 A: Let us define $n=1+3+5+7=16$.  The one digit numbers in $S$ sum to $n$.  There are $4 \cdot 3$ two digit numbers in $S$, so each digit appears in each column (ones and tens) three times. The sum of the two digit numbers is then $3(10+1)n=33n=528$.  There are $24$ three digit numbers, so each digit appears in each column six times, so the sum is $6\cdot 111 \cdot n=10656$.  Finally there are $24$ four digit numbers, so the sum is $6\cdot 1111 \cdot n=106656$.  The total is $106656+10656+528+16=117856$
A: First consider all the four digit numbers formed by these.
Digit 1 will occur in the 1000th place value for 6 numbers (when we permute 3,5,7 in the other three places, the numbers are 1357, 1375, 1735, 1753, 1573, 1537), 100th place value for 6 numbers, 10th place for 6 numbers and units place for 6 numbers. Similar for the other digits. Thus the total is $6 \times (1+3+5+7) \times (1000+100+10+1) = 106656$.
Similar argument for three digit, two digit and 1 digit numbers. 
A: You deal with the cases individually, noting that S=1+3+5+7 makes 16.
For numbers starting with 1, we note there are 3*2*1 four-digit numbers, 3*2 three-digit numbers, 3 2-digit numbers and 1 one-digit number, or 6+6+3+1 = 16 numbers.
Since this is true for each member of S individually, there are thus 64 such numbers.
We note that the position that the digits are equally spread in the digits, and that the average is then S/4 per digit, ie 4
We find that there are 24 four-digit numbers, 24 three-digit numbers, 12 two-digit numbers, and 4 one-digit number, giving 12(2222+222+11)+4 12*2455 + 4, or 117856, all together.
