I need help with combinatorics problem. The task is this: There are 9 numbers which are: 1,3,5,2,4,6,8,10,12. I need to group these numbers in 3 sets with 3 elements in every set, but there is one condition, the sum of the elements in every group needs to be odd.

I don't know what should i use, i think permutation but i don't know how to set up the answer!

Thanks very much!

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    $\begingroup$ There are three odd numbers in total, so each group must contain an odd. $\endgroup$ – Hongyi Huang May 25 at 14:27
  • $\begingroup$ Do you only have to group (i.e. find an example), or are you also asked to count the number of possible groups? $\endgroup$ – drhab May 25 at 14:30
  • $\begingroup$ So (3 1) * (6 2)? There are 3 odds and 6 evens. $\endgroup$ – Blaze May 25 at 14:30
  • $\begingroup$ Only to group these numbers in 3 groups with 3 elements in every group @drhab $\endgroup$ – Blaze May 25 at 14:31
  • $\begingroup$ Well, then I do not really see the connection with combinatorics. An example are the groups $\{1,2,4\},\{3,6,8\},\{5,10,12\}$. Just a bit of puzzling. $\endgroup$ – drhab May 25 at 14:35

Note that there are three odd numbers, so each group must have exactly one odd number, or else some group would have only even numbers and therefore have an even sum. We now need to assign two even numbers to each odd number, and this can be done by ordering six even numbers in any order, and splitting them into three groups of two, then dividing by $8$ since the order in the set does not matter. So we obtain $\frac{6!}{2^3}$ possible groupings.

  • 2
    $\begingroup$ I think ${6\choose 2}{4\choose 2}{2\choose 2}$ is better to discribe, though they are same. $\endgroup$ – Hongyi Huang May 25 at 14:33
  • $\begingroup$ Thank you very much!! And also one more question. What if in one of the three groups the sum is odd and the other two groups the sum is even? $\endgroup$ – Blaze May 25 at 14:35
  • $\begingroup$ @HongyiHuang thank you for your answer as well. $\endgroup$ – Blaze May 25 at 14:38

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