We need to find from $\{1,2,\ldots,n\}$ vertices how many graphs exist which are having exactly two cycles when self loops are allowed.
When $n=1$ there is no such graph with two cycles.
0 graph.
When $n=2$
1 graph exists.
When $n=3$
3 graphs exist.
When $n=4$
11 graphs exist.
For $n=4$
(A) $\sum_{k=1}^{n-1}k!(n-k)! = 16$
(B) $\frac{n!}{2}[\sum_{k=1}^{n-1}\frac{1}{k(n-k)}] = 11$
(C) $n![\sum_{k=1}^{n-1}\frac{1}{k}] = 44$
(D) $\frac{n!(n-1)}{2} = 36$
Option B is answer.