Find the probability that two randomly selected subsets of $\{1,2,3,4,5\}$ have exactly 2 elements common in their intersection Clearly, the total number of subsets possible is $2^5$
For two elements to be common, both subsets need to have at least two elements, so we can form quite a lot of cases which satisfy both conditions.
Now there are far too many cases (IMO) for me to manually curate, so is their a shorter way, complementary probability perhaps?
Even in complimentary method there seem to be a lot of subsets, but here is what I got
$$\binom 51(\binom 41 +\binom 42 +\binom 43 +\binom 44) + \binom 52 (\binom 31 + \binom 32 +\binom 33).....+\binom 54 (\binom 11)$$
$$=5(2^4-1) +10 (2^3-1) +10(2^2-1) +5 (2-1)$$
$$=180$$
Which comes out to be greater than $2^5$, which isn’t possible. Where am I going wrong?
Edit: My reasoning
I found the cases where one set has an element, and then the other set does not have that element, although I now realise I missed the cases where they have 1 element common or 3 or 4 or 5, which effectively makes my attempt useless.
 A: Note: $2^5$ is the count of whether each element is in a subset or not.  However, you are selecting 2 such subsets.  So we must count whether each element is included in each subset: which is $4^5$ outcomes in the space.
For the favored case, we wish a selection of 2 from the five elements to be in both, and each of the remaining 3 elements have but 3 options (in one, in the other, or in neither).  So that is $\binom 52 3^3$ favored outcomes.
As all outcomes are unbiased, the probability is: $$\dbinom 5 2\cdot \dfrac{3^3}{4^5} =\dfrac{135}{512}$$
A: There are $\binom52=10$ two-element subsets of $\{1,2,3,4,5\}$, and every pair $\langle A,B\rangle$ of subsets of $\{1,2,3,4,5\}$ whose intersection has exactly $2$ elements has one of them as its intersection. $A\cap B=S$ if and only if $A\setminus S$ and $B\setminus S$ are disjoint subsets of $\{1,2,3,4,5\}\setminus S$, so to count the pairs $\langle A,B\rangle$ such that $A\cap B=S$, we need to count the pairs of disjoint subsets of a $3$-element set.
The number of pairs $\langle C,D\rangle$ of disjoint subsets of a set $X$ of $n$ elements can be computed as follows. For $k=0,\ldots,n$ there are $\binom{n}k$ ways to choose $C$ of cardinality $k$, and there are then $2^{n-k}$ ways to choose $D$ disjoint from $C$, so by the binomial theorem there are altogether
$$\sum_{k=0}^n\binom{n}k2^{n-k}=\sum_{k=0}^n\binom{n}k1^k\cdot2^{n-k}=(1+2)^n=3^n$$
such pairs.

Alternatively, and even more simply, each of the $n$ elements of $X$ can be assigned arbitrarily to $C$, $D$, or $X\setminus(C\cup D)$, and this assignment can be made in $3^n$ ways.

In our case $|X|=|\{1,2,3,4,5\}\setminus S|=3$, so there are $3^3=27$ pairs of disjoint subsets of $\{1,2,3,4,5\}\setminus S$. Thus, there are $10\cdot27=270$ ordered pairs of subsets of $\{1,2,3,4,5\}$ that have $2$-element intersections. If you want unordered pairs, you need only cut this number in half.
A: Let's say instead of choosing two subsets from a single set, we choose subsets from the two sets $A=\{1,2,3,4,5\}$ and $B=\{1,2,3,4,5\}$, so number of subsets that can be chosen from $A$ are $2^{5}$ and similarly total number of subsets chosen from $B$ are $2^5$.
So, in a total of $2^5 \cdot 2^5=2^{10}$ ways, we can choose the subsets from $A$ and $B$.
Now, we need to find the cases, where there are exactly $2$ elements common to them.
If there are two elements in subset choosen from $A$, which can be done in $\displaystyle \binom{5}{2}$ ways, for two elements to be common, we can choose subsets in $B$ in $\displaystyle \binom{2}{2}$ ways, and for the rest of $3$ the elements, they can be chosen in $2^3$ ways.
If there are three elements in subset choosen from $A$, which can be done in $\displaystyle \binom{5}{3}$ ways, so choose $2$ elements from those three choosen for $A$, that is, $\displaystyle\binom{3}{2}$ and for rest choose them in $2^2$ ways.
If there four elements in subset choosen from $A$, there are $\displaystyle \binom{5}{4}$ ways. Choose two from those $4$, i.e $\displaystyle\binom{4}{2}$, choose rest in $2^1$ ways.
If all five elements are choosen in $A$, choose any two elements in $B$, for which there are $\displaystyle\binom{5}{2}$ ways.
So we have, probability that there is exactly one element common is
$$\dfrac{\binom{5}{2}\binom{2}{2}2^3+\binom{5}{3}\binom{3}{2}2^2+\binom{5}{4}\binom{4}{2}2^1+\binom{5}{5}\binom{5}{2}2^0}{2^{10}}=\dfrac{135}{512}$$
