# how many integers between $1000$ and $9999$ is the sum of digits equal $11$ [closed]

I have already known that all cases is $$\binom{13}{3}$$, but I don' know how to handle the bad cases, such like putting $$10$$ objects in the first box.

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• Do you know about generating functions? – Toby Mak Apr 22 at 5:43
• yes, but I dont know how to find the bad cases. – HYN Apr 22 at 5:54
• Welcome to MathSE. You should explain how you showed that there are $\binom{13}{3}$ cases in which the digit sum is $11$ before those cases in which a digit larger than $9$ is excluded. The way you wrote your answer, I initially thought you were claiming that there were $133$ such cases. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Apr 22 at 9:09

## 1 Answer

HINT

The numbers involved are simple enough that you can just manually count and exclude all cases where there are $$10$$ or $$11$$ balls in one box. (It is simple mainly because you cannot have two or more boxes each with $$10$$ or $$11$$ balls, so you don't need inclusion-exclusion.)

Thanks to @N.F.Taussig for pointing out some subtlety I didn't realize. Since the first digit is non-zero (first box has a ball), there is only $$1$$ way to put $$11$$ balls in one box. And to have $$10$$ balls in one box, either that's the first box (and the last ball can be in any other box), or the first box contains $$1$$ ball and the $$10$$ balls are together in some other box.

• so how many cases where 10 balls in one box and how many cases where 11 balls in one box. Thanks – HYN Apr 22 at 13:03
• (A) How many cases where $11$ balls in one box? There are $11$ balls, they all go in one box, and there are $4$ boxes, so... (B) If $10$ balls in one box, then the leftover one is in another box, so... – antkam Apr 22 at 13:24
• Be careful. Since the leading digit must be at least $1$, it is only possible to place $11$ balls in the first box. The remaining boxes can have at most $10$. – N. F. Taussig Apr 22 at 19:11