# 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.

• Should i be more elaborate in my answer? – Akshaj Bansal Sep 15 '19 at 14:46
• @AkshajBansal that would help a lot especially if you consider all digits in there! – Emad Sep 15 '19 at 14:50
• 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? – Akshaj Bansal Sep 15 '19 at 14:52
• @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. – Emad Sep 15 '19 at 14:53
• Repetition not allowed ,right? – Akshaj Bansal Sep 15 '19 at 14:54

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 $$45*9*8*7*1111- 45*8*7*111 = 45*8*7(9999-111)=45*8*7*9888$$

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

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 .

$9×8×7(0)(111)+8×8×7(1+2+3+4+5+6+7+8+9)(111)$\$

$$\big(1000(1+2+3+4+5+6+7+8+9)×{9\choose3}×3!\big)+ \big(8×8×7(1+2+3+4+5+6+7+8+9)(111)\big)$$ $$=1000×9×8×7×45+8×8×7×45×111=24917760$$

$$A$$ is $$8×8×7$$ and $$B$$ is $$9×8×7$$
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.
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.