just can't seem to figure this out.

I need to aquire a function for this scenario. I have tried to look at smaller forms of the problem. My problem is I am struggling to get the # of possibilities. So far I have 1:1 2:1 3:3 4:3 5:12 6:15 how do I proceed?


closed as off-topic by abel, Daniel, user147263, user99914, TravisJ May 15 '15 at 2:58

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  • $\begingroup$ Welcome to Math.SE! What are your thoughts so far? $\endgroup$ – Peter Woolfitt May 15 '15 at 2:22
  • 1
    $\begingroup$ I noticed you tagged permutations...have you tried thinking in terms of combinations? $\endgroup$ – Jared May 15 '15 at 2:30
  • 2
    $\begingroup$ Do you mean 12 groups with 2 persons in each? $\endgroup$ – HEKTO May 15 '15 at 2:30
  • $\begingroup$ Why is my question on hold? $\endgroup$ – Garret Lloyd May 15 '15 at 3:48
  • 1
    $\begingroup$ Please explain what you have "so far" a little more clearly. The remarks 1:1 2:1 3:3, etc. do not make sense. $\endgroup$ – hardmath May 15 '15 at 3:58

Here's how you approach this kind of problem when you are stuck: you look for an easier problem of the same sort, and solve that instead, and see if you learn anything that might help you with the real problem.

For example, how many ways are there to divide 2 people into groups of 2? Obviously only 1 way. That wasn't much help.

So try a harder one. How many ways are there to divide 4 people into groups of 2? Say the people are A, B, C, D. A must be matched with someone, and there are 3 people she could be matched with. Then the two unmatched people must be matched to each other, so the answer is 3.

Now how many ways are there to divide 6 people into groups of 2? Again A must be matched with someone, and there are 5 people she could be matched with, and then you are left with 4 people, and we know from the previous paragraph that 4 people can be matched in 3 ways, so the answer is $5\cdot3 = 15$.

Now you try it from there.

  • $\begingroup$ Let me make sure I understand. 7 people would be (6)(12) $\endgroup$ – Garret Lloyd May 15 '15 at 2:52
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    $\begingroup$ 7 people can't be divided into groups of 2. $\endgroup$ – MJD May 15 '15 at 3:03
  • $\begingroup$ I think I did what you said and I got 1.585times 10 to the 21 $\endgroup$ – Garret Lloyd May 15 '15 at 3:23
  • $\begingroup$ Assuming MJD's answer is $(n - 1)!!$, I got a value of $316,234,143,225$ $\endgroup$ – Jared May 15 '15 at 5:32

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