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.

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
$$\dfrac{\binom{5}{2}\binom{2}{2}2^3+\binom{5}{3}\binom{3}{2}2^2+\binom{5}{4}\binom{4}{2}2^1+\binom{5}{5}\binom{5}{2}2^0}{2^{10}}=\dfrac{135}{512}$$
