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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    $\begingroup$ Do you know about generating functions? $\endgroup$ – Toby Mak Apr 22 at 5:43
  • $\begingroup$ yes, but I dont know how to find the bad cases. $\endgroup$ – HYN Apr 22 at 5:54
  • $\begingroup$ 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. $\endgroup$ – N. F. Taussig Apr 22 at 9:09


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.

  • $\begingroup$ so how many cases where 10 balls in one box and how many cases where 11 balls in one box. Thanks $\endgroup$ – HYN Apr 22 at 13:03
  • $\begingroup$ (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... $\endgroup$ – antkam Apr 22 at 13:24
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    $\begingroup$ 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 $10$. $\endgroup$ – N. F. Taussig Apr 22 at 19:11

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