# John along with his wife,parents and four children of which two are boys and two are girls,put up at a rest house with four bed rooms for a night.

John along with his wife,parents and four children of which two are boys and two are girls,put up at a rest house with four bed rooms for a night.There are beds for two persons in each room.If either a couple or two persons of the same sex sleep in a room,what is the chance that Mr and Mrs John will sleep together?

I tried to solve the problem.
Chance that Mr and Mrs John will sleep together=favorable conditions/total conditions=2/total no of ways.

Here,i put 2,because there are 2 couples.But the total number of ways is elusive.Its answer is 1/37.

There are 4 males and 4 females.

If arranged by gender, there are ${4\choose2}{4\choose2} = 36$ ways to make 4 groups

If John and wife sleep together, so do his parents, sons and daughters, which means just $1$ way

$$\text{Thus}\;\;Pr = \frac{1}{37}$$

Edit

User264781's comments are valid, I got carried away by the "matching" answer !

The number of genderwise groups will be $3*3 = 9$ and Pr = $1/10$

• You seem to say there are $\binom{4}{2}$ ways of choosing how the males are paired up. But this is the number of ways of choosing a pair of males, and double counts the no. of room arrangements. I.e. label the males $m_1,\dots,m_4$. Then the pair $\{m_1,m_2\}$ leads to $m_1$ sharing with $m_2$ and $m_3$ sharing with $m_4$. But so does the pair $\{m_3, m_4\}$. – user264781 Sep 23 '15 at 14:52

Are you sure of the $1/37$ answer?

If John and his wife sleep together, then his parents must also sleep together. Thus their sons sleep in the same room and so do his daughters. Thus there is only one way of doing this.

If instead John doesn't sleep with his wife, then his parents also can't sleep together. Thus the rooms are all paired by gender. There are 4 males. There three people John can room with, and once this is chosen, we have determined the sleeping arrangements for the men: there are 3 ways of doing this. Similarly there are 3 sleeping arrangements for the females, and so 9 ways of John not sleeping with his wife.

So I get a 1/(9+1)=1/10 chance of John sleeping with his wife.

• yes answer is surely $\frac{1}{37}$.As per given in my book. – Brahmagupta Sep 23 '15 at 9:57
• But sometimes books make mistakes... – user264781 Sep 23 '15 at 9:58