The average number of handshakes between people around a circle. 
$5$ people are around a circle. At a moment and at the same time each
person extend their hand to one of their adjacent person. people who
extended their hand to each other, shaking their hands. In average
between different possibilities how many hand shakings are done?
$1)\frac32\quad\quad\quad\quad\quad2)1\quad\quad\quad\quad\quad3)\frac23\quad\quad\quad\quad\quad4)\frac54\quad\quad\quad\quad\quad5)\frac12$

First I noticed that for $5$ people we have $2^5=32$ different ways of extending hands (from this we can see third choice $\frac23$ can't be the answer). also maximum number of handshakes is $2$ and it happens in five different cases (at each case there is a person who is not shaking a hand). also minimum number of hand shaking is zero which happens in two cases (all persons extend their hand in a clockwise or counterclockwise directions). so I conclude there are $32-5-2=25$ cases with only one handshake hence the average is $\cfrac{2\times5+25\times1}{32}=\cfrac{35}{32}$. but I don't have this in the choices.

Establish five random variables $X_i$ for $i = 1, \ldots, 5$ where
$$
X_i = \cases{\frac12 & if person $i$ shook hands with somebody\\
0 & otherwise}
$$
Then note that the quantity we're interested in is $X_1 + X_2 + X_3 + X_4 + X_5$. Or rather, we're after the expected value of this sum.
It is known that expectation is linear. Which is to say, we have
$$
E(X_1 + X_2 + X_3 + X_4 + X_5) = E(X_1) + E(X_2) + E(X_3) + E(X_4) + E(X_5)
$$
Each of these five expectations equals $\frac 1{4}$, as there is a $\frac12$ probability that the person they extend their hand to extends back at them.
Which is to say, the final answer is $\frac54$.
As for your approach, there are acutally 10 ways to have two handshakes, as the lone person can reach out in two directions. So you get
$$
\frac{2\times 10 + 20}{32} = \frac{40}{32} = \frac54
$$
Your approach might ultimately be faster, but it is, in my opinion, easier to miss something and get the wrong answer.
