What is the sum of all 4 digit numbers with distinct digits? I tried solving this problem and got into a little trouble.I had done questions similar to this one before where the digits(that I could use) were specified so I thought maybe I could solve it like them. so I started writing all 4 digit numbers with distinct digits using the digits:  $${0,1,2,3}$$ 
and I found the number of occurrence of each digit in units place, then the 10's place ,100's place and finally the 1000's place and I got this as the sum of all numbers : $$4\times(0+1+2+3)\times(1+10+100) + 6 \times (1+2+3) \times 1000$$
and with some more thought into it I found out that I could Write the sum of the numbers in the form $$(A\times(0+1+...+9) \times 111)+(B\times (1+2+...+9)\times 1000)$$
now my question arises I'm able to find the value of B which is just $9 \times9\times8\times7$ devided by $9$ which is $9\times8\times7$ but I really can't find the Value of A because it seems to have no relations with  the value of $9 \times9\times8\times7$ so I could use some help here; or maybe what I've done is not the best way of doing it. I will appreciate your answers.
 A: Think like this 
For the sum of all digits at thousands place the expression would be
$1000(1+2+3+4+5+6+7+8+9)×{9\choose3}×3!.$
Explanation-
You surely dont have $0$ therefore now only $9$ possibilities left for thousands place suppose you take $1$ at thousands place now you need to see how many times does it repeat i.e $9\choose3$ since you need to select $3$ digits out of $9$(0,2,3...9) possibilites also you need to see in how many ways those $3$ would be arranged that is $3!$
Now-
For hundreds, tens, units place there would be two separate cases for $0$ coming at those places and ${1,2,3...9}$ coming at those places.
For $0$ we see it repeats $9×8×7$ times (but it doesnt affect the sum)
For digits ${1,2,...9}$ they repeat $8×8×7$ times ,now how does that come since you dont have $0$ at thousanda place so only $8$ possibilities and for other place $8$ possibilities again (since you can put $0$) now and $7$ possibilities for the last place .
$9×8×7(0)(111)+8×8×7(1+2+3+4+5+6+7+8+9)(111)$$
And the final answer-
$\big(1000(1+2+3+4+5+6+7+8+9)×{9\choose3}×3!\big)+
\big(8×8×7(1+2+3+4+5+6+7+8+9)(111)\big)$
$=1000×9×8×7×45+8×8×7×45×111=24917760$
Or According to your question
$A$ is $8×8×7$ and $B$ is $9×8×7$
Thank you
A: You've done well so far.  You're getting stuck where you assign a digit to one of the places other than the $1000$'s place.  Note that if you assign $0$, it contributes nothing to the sum, so you can ignore that possibility.
Once you've assigned a non-zero digit, you have to assign $3$ more, but you can't assign $0$ to the $1000$'s place.  So, calculate the number of ways to assign the three digits, without regard to the restriction, and then subtract the number of ways to assign $0$ to the $1000$'s place. 
EDIT 
There are $9$ ways to assign a non-zero digit to a place other than the $1000$'s place.  That leaves you with $9$ digits to assign to $3$ places, which can be done in $9\cdot8\cdot7$ ways.  We have to subtract the $8\cdot7$ ways where we assigned $0$ to the $1000$'s place, giving $9\cdot8\cdot7-8\cdot7=8\cdot8\cdot7=448$ ways.  
A: Take the $k$th column place.  For every digit $a$ in the $k$th columns there should be  $9*8*7$ digits in the remaining three columns.  $a$ appears in the $k$th column $9*8*7$ times.  So if you add up all the terms in the $k$ths column you should get $9*8*7(0+1+2+..... + 9) = 45*9*8*7$ and so the $k$th column provides $45*9*8*7*10^{k-1}$ to the sum.  So the total sum should be $45*9*8*7(10^3 + 10^2 + 10 + 1) = 45*9*8*7*1111$.
Except that adds up "four digit" numbers that have $0$ in the thousands column which should not be included.
So.... we subtract the sum of all strings that start with $0$.
Now by the same reasoning, if $a$ is in the $k$th column there are $8*7$  options for the remaining two columns (they can't be zero) so the sum of all those strings is $8*7*(1+2+3+ ..... + 9)*(10^2 + 10 + 1)= 45*8*7*111$.
So our sum is $  45*9*8*7*1111- 45*8*7*111 = 45*8*7(9999-111)=45*8*7*9888$
