Find the sum of all 4 digit numbers which are formed by the digits 1,2,5,6? 
I have researched and found 2 approaches but haven't understood both.Can anyone explain it clearly or probably with any real world example?

Approach 1 
Four digit numbers $ = 4 \cdot 3 \cdot 2 \cdot 1 = 24$ ways we can form a four digit number. Since it's a 4 digit number, each digit will appear $6= 24/4$ times in each of units, tens, hundreds, and thousands place. Therefore, the sum of digits in the units place is $6(1+2+5+6)=84$.  
Why they are adding $1+2+5+6$?
Similarly sum of digits at tens,hundreds,thousands place is $84$
Required sum of all numbers $= 84+ 84 \cdot 10+ 84 \cdot 100+ 84 \cdot 1000=93324$
Approach 2
I found it here:
Find the sum of all 4-digit numbers formed by using digits 0, 2, 3, 5 and 8?
Though I have understood the 2 approaches, I am not able to understand the question. 
 A: In this answer I presume that all digits $1,2,5,6$ must be present in the number.
There are $4!=24$ such numbers. 
We can index them with $i=1,\dots,24$
and write each of them as $$n_{i}=a_{i}+10b_{i}+100c_{i}+1000d_{i}$$
where $\left\{ a_{i},b_{i},c_{i},d_{i}\right\}=\{1,2,5,6\} $
Then $$\sum_{i=1}^{24}n_{i}=\sum_{i=1}^{24}a_{i}+10\sum_{i=1}^{24}b_{i}+100\sum_{i=1}^{24}c_{i}+1000\sum_{i=1}^{24}d_{i}$$
$\left\{ i\mid a_{i}=1\right\} $, $\left\{ i\mid a_{i}=2\right\} $,
$\left\{ i\mid a_{i}=5\right\} $ and $\left\{ i\mid a_{i}=6\right\} $
have equal cardinality $\frac{24}4=6$ so: $$\sum_{i=1}^{24}a_{i}=6.1+6.2+6.5+6.6=6(1+2+5+6)=84$$
This can also be applied for $b,c$ and $d$ and finally we find:
$$\sum_{i=1}^{24}n_{i}=84+10.84+100.84+1000.84=1111.84=93324$$
Does this really answers your question? If not, then please let me know.
A: For getting the sum of 4 digits numbers we can apply the formula as follows:-
$(\text{sum of given digits}) \times 1~1~1~1\cdots ~n ~(\text{given number of digits}) \times (\text{number of digits}-1)!$
eg:- to find the sum of $~4~$ digits using $~2 , ~3 , ~4 , ~5~$
sol:-  $$(2+3+4+5) \times (1111)\times ( 4-1 )!$$
$$= (14 ) \times ( 1111 ) \times ( 3! )$$
$$  = 14 \times 1111 \times 6$$
$$= 93324$$
A: We can solve by 
Average of the 'smallest number formed by all digits' and 'biggest number formed by all digits'} * {Number of given digits}!
i.e for 0,2,3,5,8
Smallest number formed by using all these digits = 20358
Biggest number formed by using all these digits = 85320
Average = (20358+85320)/2
        =  52839
So Answer to the given question is = 52839*5!
                                   = 6340680
A: for lets say 4 digit : a,b,c and d : to arrange in 4 places without repetition is 4!
So each digit appears 24/4=6 times in a position - 1st, 10th, 100th and 1000th position.
So sum of all digits in each position is same = 6*(a+b+c+d).
let say the sum is a two digit number say ef. So now the last digit of 1st position is f. The sum of 10th position is ef again but with the carry from 1st position e. Similarly sum of 100th position is ef+carry of 10th position. And so on for 1000th position.
