If $4$ people are chosen out of $6$ married couples, what is the chance that exactly one married couple is among the $4$ people? 
$6$ married couples are standing in a room. If $4$ people are chosen at random, what is the chance that exactly one married couple is among the $4$ people?


Total number of ways of selecting $4$ people out of $12$ people is $\binom{12}{4}=\frac{12!}{8!4!}=33\times 15$
I am having difficulty in counting the favourable cases.
 A: Since there seems to be some confusion among the other answers, I'll expand my comment into an answer.
You can specify each such group of four people by choosing which three of the six couples are to be represented, then which one of the three couples is to appear in its entirety, then which one of the two people from the second couple is included, and finally which one of the two people from the third couple is included.
This makes for $${6\choose 3}\cdot{3\choose 1}\cdot{2\choose 1}\cdot{2\choose 1}=20\cdot3\cdot2\cdot2$$
favorable outcomes. Dividing by the ${12\choose4}=33\cdot15$ total outcomes gives $16/33$.
A: Answer:$$\frac{6.\binom52.2^2}{\binom{12}4}=\frac{16}{33}$$
Explanation of numerator:


*

*$6$ possibilities to choose a couple that provides $2$ selected persons.

*$\binom52$ possibilities to choose $2$ couples that provide exactly $1$ selected person.

*For the $2$ couples that provide exactly $1$ selected person there are $2$ possibilities.

A: Answer by two approaches
Using combinations
$\dfrac{\binom61\cdot\binom52\cdot2^2}{\binom{12}{4}}= \dfrac{16}{33} $
[ The $2^2$ is to account that either of the couple may be chosen ]
Using permutations
First pair can be chose and lined up in $6\cdot4\cdot3$ ways, and the rest of the numerator takes care that no more pair is selected.
$Pr = \dfrac{72\cdot10\cdot8}{12\cdot11\cdot10\cdot9} = \dfrac{16}{33}$ 
A: See we will select a couple so it can be selected in $6$ ways  now we are left with $5,5$ men,women so we shouod now make cases . Selecting both men or both women or one man one woman so the ways of selecting men ${5\choose 2}=10$,selecting women =$10 $ now selecting one man,one woman but from different couples so ways here are ${5\choose 1}{4\choose 1}=20$ so total ways are $6(10+10+20)=240$ so probability becomes  $\frac{240}{495}=\frac{16}{33}$ 
