CODEWITHSUNDEEP
randomsplitcombinationspermutationpartitioning
wishiknew
wishiknew

Reputation: 1

How many combinations of partitions of a list can you make so that repetitions are minimised?

Well one line is not enough to capture that question so ill explain it here.

Earlier today i was in a zoom meet where there are 12 members and we were split into different breakout rooms. After 30 mins we were shuffled and the host made sure to use a different combinations such that no two people who were in the same breakout room was in the same breakout room now. The host atleast tried to minimize the repetitions.

So my question is how many such combinations can we make on a group such that no two people who were in a meeting earlier repeat themselves.

If you could tell me in a algorithmic way, i will understand it.

Edit: Let me further explain my problem with a use case:

Suppose there are 12 people in a zoom meet who are to be divided up into breakout rooms. Let me form 4 breakout rooms with 3 in each group.

so let the combination be [1,2,3][4,5,6][7,8,9][10,11,12]. the first iteration could have been anything really. now the problem starts. for the next combination i have to make sure that the breakout rooms cannot have the same people forming it. so keeping this in mind i'll form the group as: [1,5,7][4,8,12][2,6,11][3,9,10]. wow. did you just see that? i had to make sure that no two people who were in a group earlier, are present in a single group presently. So my question is how do i write an algorithm for this problem.

Upvotes: 0

Views: 109

Answers (1)

Paul Pascher
Paul Pascher

Reputation: 317

math related questions fit better in https://math.stackexchange.com/

to your question:
i assume breakout rooms mean pears of 2?
-> everyone can match with 11 others - so basically 11 rounds? There are no conflicts necessarily that reduce the rounds, if you organized matches from the start on.

Edit:
If you instead looked for the commonly school question:

Suppose there is a group of 12 people. In how many different ways can the 12 people be split into six pairs?

Then there are same matches allowed as long as one pair changes and the result would be
12!/(2^6 x 6!)

see: here

Edit after Clarification:

Well I only know Primary math and would still recommend https://math.stackexchange.com/ : )

My unqualified guess would be you have your total Permutations of 12! / (12-3)! Which are only 12! /(3! *(12-3)!) Combinations and then reduce it by all non-valid Combinations?

   ( 12! / (3! *(12-3)!) ) / 4 = 55?
Or   12! / (4! *(12-3)!)       = 55?

If you need this for some kind of school, I recommend learning basics first. The solution will not help you in your class!

https://www.calculator.net/permutation-and-combination-calculator.html?cnv=12&crv=3&x=76&y=12 https://math.stackexchange.com/questions/35684/combination-of-splitting-elements-into-pairs

Upvotes: 1

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