# Number between 1 and 1000000 with sum of digits equal to 18.

I saw an example of such a problem here already, where the sum of digits desired was 8. I have already done my question using multinomial theorem, by finding out the coefficient of x^18 in $$(x^0+x^1+...+x^9)^6$$, and the proper answer came (which is 25927). How to solve it using the fact that $$x_1+x_2+...+x_6=18$$, and hence through $$18+5\choose 5$$? I observed the fact that I get the term $$18+5\choose 5$$ in the first way also, but some options get eliminated after that (subtracted). I understand that those are the cases where the digits are something greater than 9, but is there a proper mathematical explanation for it? Thanks in advance.

Well, I think you gave the explanation yourself already: the $$18+5\choose 5$$ is for any $$6$$ (ordered!) numbers that add up to $$18$$, and that includes more cases than having $$6$$ digits that add up to $$18$$ .. I am not sure what else you might be looking for in order to explain this.

But maybe this helps ....

In terms of stars and bars: finding the ways in which $$6$$ numbers add up to $$18$$, you'd need to put $$5$$ bars between or on the sides of $$18$$ stars, e.g

$$****|**||*|***********|$$

corresponds to the ordering of $$6$$ numbers $$(4,2,0,1,11,0)$$, and you can see there are $$18+5\choose 5$$ ways to put those $$5$$ bars in those $$18+5=23$$ positions

However, when it comes to $$6$$-digit numbers (where a number like $$53$$ is represented by $$000053$$), you can't put the bars more than $$9$$ spots apart, and so there are fewer ways to do that.

• Thank you. I think I was confused with my own explanation (if that even means anything logical). The last paragraph was what I was missing. – Arka Seth Apr 3 '19 at 17:34