Math Is Fun Forum

goed

Member

Registered: 2008-02-05

Posts: 4

Arranging people in a circular table

A meeting involves 4companies, each company sends three representatives- MD,CA,CS.In how many ways can the twelve people can be arranged in a circular table if the three people from each company  sit together with MD in between the CS and CA in each case??

This is how I tried....but couldn't get the right answer(96)

Step 1: There are 4 companies sending 3people and all the three people should sit together so there are 4 sets which can be arranged in 4! ways i.e 24 ways
Step 2: MD should alwys be between CS and CA so CS and CS can be arranged in 2! ways i.e 2ways

Hence the total number of ways  is 24x2= 48 ways.

But i know i didn't arrange them in a circular table....how can we do that??

Thanks

NullRoot

Member

Registered: 2007-11-19

Posts: 162

Re: Arranging people in a circular table

This is my guess. Feels a bit... incomplete though.

There must be 12 chairs here so that 4x3 people can sit down. We know that there must be two empty seats between each MD because they each have to be flanked by CA and CS. That means there are 12 places to seat MD1 and the other three are forced into their positions. 

For each of those 12 options, the other three MDs have 6 ways to arrange themselves (423, 432, 234, 243, 342, 324).

So that's 72 configurations, I believe? Now for each of those 72 configurations, the CSs and CAs can be in two configurations per MD (CS, MD, CA or CA, MD, CS).

If we represent this configurations as a 4-digit number with 1 representing CS/MD/CA and 2 representing CA/MD/CS. So 1111, for instance, would represent all four companies in the CS/MD/CA configuration and 1122 would mean the first two are in CS/MD/CA and the second two are in CA/MD/CS.

We could have 16 configurations then (1111, 1112, 1121, 1122, 1211, 1212, 1221, 1222, and the same 8 again with 2 as the first digit).

So I would make it 72x16 configurations or 1216 configurations.

Still not sure though. It'd be nice to get a second opinion.

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Trillian: Five to one against and falling. Four to one against and falling… Three to one, two, one. Probability factor of one to one. We have normality. I repeat, we have normality. Anything you still can’t cope with is therefore your own problem.