[Math] A committee of three people is to be chosen from four married couples.

combinations

A committee of three people is to be chosen from four married couples.

How many committees are there such that no two members of the
committee are married to each other?

Best Answer

To avoid choosing both persons of a couple, split the process in two steps:

  1. Choose $3$ couples out of the $4$ couples. This can be done in $\dbinom{4}{3}=4$ ways.
  2. From each of these $3$ couples, choose $1$ person. You can do this in $\dbinom{2}{1}\dbinom{2}{1}\dbinom{2}{1}=2^3=8$ ways.

By the multiplication principle (or rule of product) this gives you $$4\cdot8=32$$ ways to complete this task (under the given constraint).

Another way to solve this exercise is the following: For the first person, you have $8$ persons to choose from. Pick one and remove its spouse. So for the second person you have $6$ persons to choose from. Pick one and remove its spouse. For the last person you are left with $4$ persons, from which you should choose $1$. This gives you $$8\cdot6\cdot4=32\cdot6$$ ways to form the committee, but be careful: as of this approach order matters. Hence, to obtain the correct result divide with $3!=6$ which gives $$\frac{32\cdot6}{3!}=32$$ as above.

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