# 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.

• Hint: this happens when the four people come from exactly three couples. (One couple in its entirety, one person from another couple, one other person from another couple.) Jan 11, 2016 at 8:35
• I got it,thanks.
– diya
Jan 11, 2016 at 8:45
• @diya It will be nice if you post your answer if you have got it. Jan 11, 2016 at 8:46

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 $16/33$.

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.

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.

$Pr = \dfrac{72\cdot10\cdot8}{12\cdot11\cdot10\cdot9} = \dfrac{16}{33}$

• The answer given is $\frac{16}{33}$
– diya
Jan 11, 2016 at 9:10
• Hey there are $12$ people in all how did you create $18$ people?? Jan 11, 2016 at 9:10
• @diya you can see my answer. Jan 11, 2016 at 9:10
• @diya: I will explain more a little later, there is a computer glitch now. Jan 11, 2016 at 9:27
• @diya: Sorry for the confusion and delay, but you now have many approaches ! Jan 11, 2016 at 9:51

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 $\frac{240}{495}=\frac{16}{33}$

• The question doesn't say that of the remaining two, one has to be a man and one a woman. Jan 11, 2016 at 9:24
• I have edited my answer see it niw. Jan 11, 2016 at 9:45
• I don't understand the logic of 10+10+20 Jan 11, 2016 at 10:00
• See they are cases 1) selecting only men 2) selecting only women 3) one man one woman Jan 11, 2016 at 10:12
• Oh, you mean one man one woman from different couples.. (+1) Jan 11, 2016 at 11:58