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Actually , my doubt is about number of ways of distributing 12 distinct balls to 5 different persons such that 3 persons get 2 balls each and 2 persons get 3 balls each.

There is another question in this site about "How many ways to divide group of 12 people into 2 groups of 3 people and 3 groups of 2 people?" But my question is slightly different.

Solutions with explanation is needed. Thank you anyways for your help!

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  • $\begingroup$ the ans should be 12!/(2!2!2!3!3!) since it is equivalent to permutation of 12 balls but dont care about the order of balls giben to any of the 5 ppl. $\endgroup$ – Ben Nov 26 '16 at 4:51
  • $\begingroup$ But, can't I say that after I've grouped the 12 balls into 3 grps of 2 balls and 2 grps of 3 balls, there are 5!/(2!3!) ways of distributing them to the 5 different persons? $\endgroup$ – Preetham Krishna Nov 26 '16 at 7:50
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I believe that the correct answer is the product of:

  • The number of ways to choose the persons: $\frac{5!}{3!2!}=10$
  • The number of ways to choose the balls: $\frac{12!}{2!2!2!3!3!}=1663200$
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  • $\begingroup$ Waiting for some comments from any combinatorics expert on this website (true blue anil, fleablood, Graham Kemp, JMoravitz, N. F. Taussig, etc)... $\endgroup$ – barak manos Nov 30 '16 at 7:12
  • $\begingroup$ thanks, I think your answer sounds right. I modified my answer. By the way, arent you a combinatorics expert? $\endgroup$ – vidyarthi Nov 30 '16 at 7:16
  • $\begingroup$ @vidyarthi: Can't testify for myself but I know that they are :) ... Not yet entirely convinced that this answer is correct, so will wait for a few more comments... $\endgroup$ – barak manos Nov 30 '16 at 7:21
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The answer is $\binom{12}{2}.\binom{10}{2}.\binom{8}{2}.\binom{6}{3}.\binom{3}{3}$. This is because, we choose 2 balls out of 12 for the first person, 2 out of 10 for the second person, 2 out of 8 for the third person. The remaining 6 balls are chosen 3 each for the remaining 2 people. The answer comes to be the same as posted in the comment of $\frac{12!}{2!2!2!3!3!}$, which is the same as permuting 12 objects in groups of 2's 3 times and 3's 2 times.


EDIT: The above answer is correct as long as you may may assume that any person may get the 3 balls and /or 2 balls. But, as Barak Manos has pointed out in his answer, the answer for the different person getting 3 or 2 balls based on his selection,i.e. for the distinctness of the people, is $\frac{5!}{2!3!}\cdot\frac{12!}{2!2!2!3!3!}$

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  • $\begingroup$ But, can't I say that after I've grouped the 12 balls into 3 grps of 2 balls and 2 grps of 3 balls, there are 5!/(2!3!) ways of distributing them to the 5 different persons? $\endgroup$ – Preetham Krishna Nov 26 '16 at 7:50
  • $\begingroup$ @PreethamKrishna While grouping you are actually choosing 2 from 12 balls, 2 from remaining 10 balls, 2 from remaining 8 balls and so on. If the groups of 3 balls and 2 balls were given beforehand, then your argument is right. But since you have to group them before distributing, your argument fails $\endgroup$ – vidyarthi Nov 26 '16 at 7:55
  • $\begingroup$ @PreethamKrishna: In my understanding, this answer assumes that the persons are indistinguishable, and is thereore wrong. I believe that it should be multiplied by $\frac{5!}{3!2!}$, but let's wait for some more comments on it. $\endgroup$ – barak manos Nov 30 '16 at 6:59

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