How many of the n! permutations π from set N to N satisfy min(π(A)) = min(π(B)) Given a set of elements $N = \{1, 2,\ldots, n\}$ and two arbitrary subsets $A\subseteq N$ and $B\subseteq N$, how many of the $n!$ permutations π from $N$ to $N$ satisfy min(π$(A)$) = min(π$(B)$), where min(S) is the smallest integer in the set of integers S, and π(S) is the set of integers obtained by applying permutation π to each element of S?
I find it difficult to approach this problem.
First I assume $n =3$, so $N=\{1,2,3\}$.
I consider $A=\{1,2\}$ and $B = \{2,3\}$.
Then apply π$(A)$, which i have doubt how to form set of integers.
Suppose π$(A)= \{1,2\}$ and π$(B)=\{1,2,3\}$.
I don't get it how to apply min(π$(A)$)=min(π$(B)$).
The options to the question are
(A) $(n - |A ∪ B|) |A| |B| $
(B) $(|A|^2+|B|^2)n^2$ 
(C) $n!(|A∩B|/|A∪B|)$ 
(D) $(|A∩B|)^2/(n/|A∪B|)$
 A: I will use $\rho$ rather than $\pi$ to represent an arbitrary permutation.
If $\text{min}(\rho(A)) = \text{min}(\rho(B))$, say $y = \rho(x)$ is the common minimum.  Since the permutation $\rho$ is a 1-to-1 mapping of $N$, it follows that $x \in A\cap B$.  So the intersection $C = A\cap B$ cannot be empty for this to happen.
The desired outcome is that $y = \rho(x) \in \rho(A\cap B)$ is the minimum of $\rho(A\cup B)$, if the minimums of $\rho(A)$ and $\rho(B)$ are to be the same.
The permutation $\rho$ must send $A\cup B$ to any equal-sized subset of $N$, and the desired outcome depends only on whether the least element of $\rho(A\cup B)$ belongs to $\rho(A\cap B)$.  Considering all possible permutations, the fraction of them that meet this condition is simply $|\rho(A\cap B)|/|\rho(A\cup B)|$.  It follows from the 1-to-1 mapping that this fraction is also $|A\cap B|/|A\cup B|$, and the total number is given by multiplying by $n!$:
$$ n! \frac{|A\cap B|}{|A\cup B|} $$
This agrees with one of your multiple choice answers.
To use your example of $N=\{1,2,3\}$ with $A=\{1,2\}$ and $B=\{2,3\}$, we see that a  permutation can act on $N=A\cup B$ in six ways.  Of these, one-third will map $2 \in A\cap B$ to the minimum position.  Therefore the total number of permutations such that $\text{min}(\rho(A)) = \text{min}(\rho(B))$ is six times one-third:
$$ 3! \frac{|A\cap B|}{|A\cup B|} = 6 \cdot \frac{1}{3} = 2 $$
A: The number of permutations you are after is $\sum_{i\in N}|\{ \,\pi\mid \min(\pi(A))=i=\min(\pi(B))\,\}|$. You can compute each of the terms in that summation fairly easily.
