Using probability properties to solve exercise. I am not able to understand whether this exercise is correct or not.
There are N teams formed by 2 people, one male, and one female from N married couples
And it is asking me to calculate the probability of


*

*No female is paired with her husband in the same team

*having exactly k teams formed by husband wife in the same team.


My solution was to start it calculating the probability of the probabilty of the total space and then divide this number by the probabilities of 1 and 2. the problem here is i don't understand how to finde these probabilities.
 A: @BGM is correct that you can use Derangements. From that, $D_{N,0}$ is the number of permutations of the set $\{1,2,\ldots,N\}$ that leave no elements in its original position.
Now, think of a bipartite graph, where the two sets represent male $M = \{1,2,\ldots,N\}$ and female $F=\{1,2,\ldots,N\}$, and there is an edge between a male and female iff they are paired. Each pairing (or matching) corresponds to a permutation on $\{1,2,\ldots,N\}$. 
As an example, let $N=3$. Then the pairing where each male is paired with his wife corresponds to the identity permutation, $1 \to 1$, $2 \to 2$ and $3 \to 3$.
On the other hand, the permutation $1 \to 2$, $2 \to 3$ and $3 \to 1$ is a derangement. In this specific case, the number of derangements is $D_{3,0} = 2$.
For the first part, assuming that all permutations are equally likely, the probability is $\frac{D_{N,0}}{N!}$, where $N!$ is the total number of permutations on the set $\{1,2,\ldots,N\}$.
For the second part, since $k$ men are paired with their respective wives, we first need to count the number of ways to choose $k$ out of $N$ men, which can be done in $\binom{N}{k}$ ways. Each of the remaining $N-k$ men should then be paired with a female other than his wife. So the probability in this case is $\frac{\binom{N}{k}D_{N-k,0}}{N!}$.
