-2
$\begingroup$

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.

$\endgroup$

closed as unclear what you're asking by GAVD, steven gregory, farruhota, John Omielan, Eevee Trainer Apr 23 at 4:17

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.