2n Clowns are sitting in a carouse. Each of them sees exactly one clown in front of him. In how many ways they can swap places so each one will see a new clown in front of him if the places are marked?
Answer:
$|U|= (2n)!$ are all the possibilities
We define sets:
A1 - group of all options from U where pair # 1 is facing each other
A2 - group of all options from U where pair # 2 is facing each other
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An - group of all options from U where pair # n is facing each other
$|\bar{A_{1}}\cap \bar{A_{2}}\cap..\cap\bar{A_{n}} |$
According to the principle of exclusion and inclusion we will get
Can I get more elaboration on the answer?
What does $\bar{A}$ mean in term of the definition of $A$ above?
What does $|\bar{A_{1}}\cap \bar{A_{2}} |$ mean in term of the definition?
What are the all options for example for group A1 that is defined above, how is calculated combinatory?
$\bar A_i$ means here the complement of $A$, i.e. group of all options from $U$ where pair #$i$ is not facing each other.
$|\bar{A_{1}}\cap \bar{A_{2}} |$ means here the number of all options from $U$ where neither pair # 1 nor pair # 2 are facing each other.
The number of all options in $A_1$ is $2n(2n-2)!$ where the factor $2n$ counts the number of ways to choose the cabin and places in the cabin for the pair #1, whereas the other clowns can choose their places arbitrarily in $(2n-2)!$ ways.
Similarly $|{A_{1}}\cap{A_{2}} |=2^2n(n-1)(2n-4)!$ and so on. The final formula is the special case application of the inclusion-exclusion principle.