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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.

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  • $\begingroup$ Should i be more elaborate in my answer? $\endgroup$ – Akshaj Bansal Sep 15 '19 at 14:46
  • $\begingroup$ @AkshajBansal that would help a lot especially if you consider all digits in there! $\endgroup$ – Emad Sep 15 '19 at 14:50
  • $\begingroup$ Do u wnt an answer when we consider $0,1,2,3,4,5,6,7,8,9$ as digits for the 4 digit numbers? $\endgroup$ – Akshaj Bansal Sep 15 '19 at 14:52
  • $\begingroup$ @AkshajBansal yes just as the title suggests, I used 4 digits so that I could write down all of the digits myself and use it to find the formula for the general case. $\endgroup$ – Emad Sep 15 '19 at 14:53
  • $\begingroup$ Repetition not allowed ,right? $\endgroup$ – Akshaj Bansal Sep 15 '19 at 14:54
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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$

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  • $\begingroup$ that is true and the answer also matches my book's.thank you! $\endgroup$ – Emad Sep 15 '19 at 16:43
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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

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  • $\begingroup$ I know that and besides I used the digits 0,1,2,3 to find a formula for the real question which is the sum of all 4 digits with distinct digits. $\endgroup$ – Emad Sep 15 '19 at 14:48
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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.

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