Interesting probability question - husband and wife committee variation Twenty husbands and wives (ten couples) are randomly divided into two groups. What is the probability that at exactly 4 wives are in the same group
as their husbands?
Attempt: There are $\binom{40}{2}=780$ numbers of posible committees by dividing the couples into two groups. The way I thought about this problem was to have a bag filled with two colored balls both numbered 1-10 and drawing them out one by one. If there are exactly 4 couples together in a group that means there must be 1 male female pair in a committee with a different partner, or with the ball example two different colored balls with different numbers. So the number of ways of having a committee with exactly 4 couples is the number of pairs $$(1,2), \dots (1,10), (2,3), \dots (2,10), (3,4), \dots (3,10), \dots (9,10) = \sum_{i=1}^9 i = 45$$
therefore the probability is $45/780.$ Can anyone confirm this answer? I think it is wrong and I also think there is a much more elegant solution.
 A: It is not clear to me what exactly is meant by "...are randomly divided into $2$ groups...". I preassume that the group of $20$ is randomly split up in a group $A$ and a group $B$ both containing $10$ persons.
If there are exactly $k$ married couples in group $A$ then automatically
there are exactly $k$ married couples in group $B$.
So exactly $4$ women in the same group as their husband requires
that $k=2$.
There are $\binom{10}{4}$ ways to choose $4$ married couples.
There are $\binom{4}{2}$ different ways to
put $2$ of those couples in group $A$ and the other $2$ in group $B$.
There are $2^{6}$ ways to choose $6$ persons from the remaining
$12$ such that among them there are no couples. The chosen $6$ are put in group $A$ and the remaining $6$ in group
$B$.
There are in total $\binom{20}{10}$ ways to choose $10$ persons
for group $A$.
Probability:$$\frac{\binom{10}{4}\binom{4}{2}2^{6}}{\binom{20}{10}}\sim0.4365$$
A: Your thinking is wrong. You're assuming that having exactly 4 couples in one group would mean there's space for 2 other people left in that group. But in that case, the partners to these two people would be in the other group, as well as everybody else - and "everybody else" is four more couples, so you would have a total of 8 couples, not 4.
The only way to have exactly 4 couples would be to have exactly 6 non-couples, so 6 people are in one group and their partners in the other. This makes it clear that you would have to have 2 couples in one group, 2 couples in the other to total 4 couples.
I'm really bad at probabilities and I also haven't done it for years, so I'll leave the math to somebody who's better at it (sorry).
A: I'm considering the following model: Each person is assigned to one of the groups $H$ or $T$ by throwing a coin.
For any given couple the probability that both members are assigned to the same group is  ${\displaystyle{1\over2}}$.
The probability $P$ that exactly four couples stay together then computes to
$$P={{10\choose 4}\over 2^{10}}={105\over512}\ .$$
A: I am assuming that there are $10$ people in each group
The number of "singles" in a group must necessarily be matched by an equal number in the other group, and thus so, too, must  the number of couples, which means that for  exactly $4$ couples,
each group must have $2$ couples.
Two couples can be chosen and lined up in $\binom{10}2\cdot10\cdot9\cdot8\cdot7 = 226,800$ ways and for the remaining, the rest of the numerator ensures that no other couple is selected
$$\begin{align}Pr = & \frac{226,800\cdot 16\cdot 14\cdot 12\cdot 10\cdot 8\cdot 6}{20\cdot19\cdot18\cdot17\cdot16\cdot15\cdot14\cdot13\cdot12\cdot11} \approx 0.4365\end{align}$$ 
A: You Could easily count the number of such partitions for a general partitions, and then dependent of your probability space you could use that to infer the probability.
We want $4$ couples that are in the same group and $6$ that are in different. So we assume that there is two different groups $A, B$ and we later divide by $2$ so that cases $(A,B) , (A',B')$ such that $A'= B$ will be the same in our point of view.
So if there are two different groups we, first we decide which couples going to be in the same group - ${10}\choose{4}$ possible selections.
For each of those couples, we decide whether it will be in $A$ or $B$ - $2^4$ possibilities.
Now finally, we decide about the other couples. since they cant be in the same group, if we decide that the husband is in $A$ it would mean that the wife must be in $B$, so we should just choose for the husbands their group - $2^6$ possibilities.
Multiplying them all we get that we have ${{10}\choose{4}} \cdot 2^{10}$ such partitions into different groups.
