This is in reference to the below question https://gateoverflow.in/473/gate2012-38 My doubt here is In this question, we can also solve like first we select 4 vertices out of 6 in 6C4 ways. Now we need to count how many cycles we can form with 4 vertices.And I think this is ... is (n-1)! /2 which evaluates to 3 in case of n=4.So by multiplication rule, total cycles we have 45.Is this way correct?

I was searching for this question and found below link How many subgraphs does $K_5$ have? Here, in general, it is given to be $\displaystyle\sum_{r=0}^{n}\binom{n}{r}2^{\frac{r(r-1)}{2}}$ My doubt is that the case for $r=0$ is taken if we select none of the vertices of $K_n$? Why the case for $ r=0$ considered?