The probability that two randomly selected subsets of the set $\{1 , 2, 3, 4 , 5\}$ have exactly two elements in their intersection , is The probability that two randomly selected subsets of the set $\{1 , 2, 3, 4 , 5\}$ have exactly two elements in their intersection is $\frac{135}{2^9}$.
In this problem Find the probability that two randomly selected subsets of $\{1,2,3,4,5\}$ have exactly 2 elements common in their intersection , it is said that the sample space has $2^{10}$ elements.
I can not understand how sample space contains $2^{5}\times 2^5$ elements. It should have $2^5 \choose 2$ elements. It has $\binom{2^5}{2} + 2^5$ elements if we allow two chosen subsets being identical.
 A: Method 1:  Let's say the sets are $A$ and $B$.  You have four choices for each of the five elements in the set $\{1, 2, 3, 4, 5\}$: place an element in both sets $A$ and $B$, place it only in set $A$, place it only in set $B$, or place it in neither subset.  Hence, the sample space has $4^5$ elements.
Method 2: There are $2^5$ ways to select set $A$ since we can choose to include or not include each element of the set $\{1, 2, 3, 4, 5\}$ in set $A$.  For each such choice, there are $2^5$ ways to select subset $B$.  Since the choice of sets $A$ and $B$ is independent, there are $2^5 \cdot 2^5 = 2^{10}$ ways to choose the two subsets of the set $\{1, 2, 3, 4, 5\}$.
What is wrong with your approach?
While you did not explain your reasoning, it appears that the term
$$\binom{2^5}{2}$$
is supposed to count the number of ways of choosing two different subsets of the set $\{1, 2, 3, 4, 5\}$, while $2^5$ is supposed to count the number of ways of choosing two identical subsets.
Method 3: There are $2^5$ ways of choosing the subset $A$ and one way to select set $B$ so that it is the same as set $A$.  Thus, the term $2^5$ makes sense.
For each of the $2^5$ ways we can select set $A$, there are $2^5 - 1$ ways to select set $B$ such that set $B$ is different from set $A$.  Hence, there are $2^5(2^5 - 1)$ ways to pick two different subsets.
Thus, we again obtain
$$2^5 + 2^5(2^5 - 1) = 2^5 + 2^{10} - 2^5 = 2^{10}$$
elements in the sample space.
Note:  The problem with using $\binom{2^5}{2}$ for the number of ways of selecting two different subsets is that there are two orders in which we could pick the same pair of distinct subsets.  Notice that
$$2\binom{2^5}{2} + 2^5 = 2 \cdot \binom{32}{2} + 2^5 = 2 \cdot \frac{32 \cdot 31}{2} + 32 = 32 \cdot 31 + 32 = 992 + 32 = 1024 = 2^{10}$$
A: Because the subsets, let's say $A$ and $B$, can be the same we must treat them as distinguishable. Hence $A$ has $2^5$ choices and $B$ has $2^5$ choices for a total of $2^{10}$ ways to pick $A$ and $B$.
If we use your result of $\binom{2^5}{2}+2^5$, you are saying that each of the cases when $A$ and $B$ are identical has twice the probability of occurring when compared to a case when $A$ and $B$ are distinct. This is clearly not the case as each possible ordered pair $A,B$ should have the same probability of occurring.
