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With how many ways can we put $12$ numbered balls in $4$ similar(not numbered) boxes of capacity $3$ each one?

Is it maybe $3^4$ ?

evinda
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1 Answers1

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You can choose $3$ out of $12$ balls for the first box, $3$ out of $9$ for the second...and so on...But since the boxes are similar you should divide the result by all the possible combinations of the $4$ boxes ($4!$), so the number requested is: $$\frac{\binom{12}{3}\cdot\binom{9}{3}\cdot\binom{6}{3}\cdot\binom{3}{3}}{4!}=\frac{\frac{12!}{9!\cdot 3!}\cdot\frac{9!}{6!\cdot3!}\cdot\frac{6!}{3!\cdot3!}\cdot\frac{3!}{3!\cdot0!}}{4!}=\frac{12!}{3!^4\cdot4!}=15400$$ If the boxes weren't similar the solution would be: $$\binom{12}{3}\cdot\binom{9}{3}\cdot\binom{6}{3}\cdot\binom{3}{3}=\frac{12!}{9!\cdot 3!}\cdot\frac{9!}{6!\cdot3!}\cdot\frac{6!}{3!\cdot3!}\cdot\frac{3!}{3!\cdot0!}=\frac{12!}{3!}=369600$$

sirfoga
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  • Althought the boxes are similar?? And how would it then be,at the case when the boxes were not similar???? – evinda Apr 26 '14 at 12:47
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    I've edited, check the edit... – sirfoga Apr 26 '14 at 12:48
  • I saw it but I haven't understood it.. Why is it like that?? – evinda Apr 26 '14 at 12:50
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    I'll add further information – sirfoga Apr 26 '14 at 12:51
  • Why do we have to divide the result by all the possible combinations of the 4 boxes?? I haven't understood it.I am confused. – evinda Apr 26 '14 at 12:57
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    Because once you filled all the boxes, it doesn't matter which is the first one, second one, and so on...since they're similar, you can exchange their places and the order according which you filled the boxes! – sirfoga Apr 26 '14 at 13:02
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    If you're not convinced, try figuring out how the solution would be with less boxes, say $2$ similar boxes of capacity $2$. The result would be $$\frac{\binom{4}{2}\cdot\binom{2}{2}}{2!}=3$$ In fact all the possibilities are $$AB,CD\qquad AC,BD \qquad AD,BC$$ – sirfoga Apr 26 '14 at 13:04
  • I understand..Thanks!!! – evinda Apr 26 '14 at 13:06
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    You're welcome! Understanding the reasoning it's always the most important thing! ;) – sirfoga Apr 26 '14 at 13:09
  • To check if I understood it,could you tell me if the answer of a similar question is right? A group of $20$ people want to create $3$, foreign to each other,committees with $6,5 \text{ and } 4$ people respectively.In a commitee the members are equal.With how many ways can that be done.I thought that the result is:

    $$\frac{\binom{20}{6} \cdot \binom{14}{5} \cdot \binom{9}{4}}{3!}$$

    Could you tell me if it is right???

     – evinda Apr 26 '14 at 14:03
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    Sorry, I've been outdoor a bit...anyway...YES! it's correct! ;) There are $6$ choices out of $20$ for the $1$st committee, $5$ out of $14$ for the second one, and $ 4$ out of $9$ for the last one. Finally there are $3!$ permutations of the $3$ committees, so the result must be divided by $3!$. – sirfoga Apr 26 '14 at 17:08
  • Great!!!Thank you very much!!! – evinda Apr 26 '14 at 17:51