A committee of three people is to be chosen from four married couples. 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?
 A: To avoid choosing both persons of a couple, split the process in two steps:


*

*Choose $3$ couples out of the $4$ couples. This can be done in $\dbinom{4}{3}=4$ ways.

*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.
A: Initially there are 8 people. You can choose any one of those. So 8 choices.
After choosing one you cannot choose her/his husband/wife. So you are left with 6 people. Now it is 6 choices. 
After choosing the second person you are left with 4 choices.
But they can be arranged in 6 ways so answer I think is 192/6=32
