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Untwinned twins

Posted by Chris on September 1, 2014 – 2:45 pm

A class has six pairs of twins. The teacher wishes to set up teams for a quiz, but doesn’t want to put any pair of twins in the same team.
1) In how many ways can they be split into two teams of six?
2) In how many ways can they be split into three teams of four?

This post is under “MathsChallenge” and has 2 respond so far.

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1. 1. jan janse Said：

the twins have to be split so the only variation in dividing in two groups of 6 is which twin is in which group. so we have 2^6=64 possibilities. But because of symmetry we have to divide this by 2 so we get 32 possibilities.

Suppose a,b,c and d are the first group, there are 12*10*8*6=5760 possibilities. That leaves two sets of twins not in a group yet. These have to be divided among the two final groups, considering this we have 4*5760=23040 possibilities. The remaining 4 people can be put in any group in any combination. So this gives 23040*(4!/(2!*2!))=23040*6=138240 possibilities.
Because of symmetry we can divide this by 3!=6. And we get 23040 total possibilities.

2. 2. Chris Said：

Hi Jan. I’m sorry I’ve taken so long to respond.

You forgot to divide 5760 by 4! to get 240 as the order of selection is irrelevant. So you end up with 960. Other than that slip, I think you gave a good analysis.

I don’t remember seeing an expression like 12*10*8*6 before. I note that may be written as 2^4 * 6*5*4*3 = 2^4 * 6! / 2!. It happens, we also want to divide that by 4! (as order is irrelevant), so we end up with 2^4 * C(6,4), and with hindsight, that’s just what we should have expected.

Using that, here’s another way to do the problems. For the two teams of six, we have 2^6 * C(6,6) /2! = 2^5. The divisor of 2! is because of the symmetry with two teams.

For the second question, we have 2^4 C(6,4) = 16 * 6!/(4! 2!) = 16 * 15 = 240 for the first team. Then we have 2^2 C(2,2) = 4 * 1 = 4 ways of putting the remaining two pairs of twins into the other teams (i.e. (ef, EF), (eF,Ef), (Ef,eF), (EF,ef)). And finally C(4,2) = 4!/(2! 2!) = 6 ways of distributing the remaining people (e.g. A,B,C and D),

Again, symmetry requires dividing by 3! = 6. So the grand total is 240*4*6/6 = 960.

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