So starting with minimum number of edges.
Given ,
Number of edges =$m$
Number of vertices=$n$
Number of Component=$k$
Now say each component $G_{1},G_{2},G_{3},.....G_{k}$ has $n_{1},n_{2},n_{3},.....n_{k}$ vertices respectively.
So , $n_{1}+n_{2}+n_{3}+....+n_{k}=n$
each component is a connected graph. So, we have $k$ such connected graph as it is given we have k connected component.
Now consider Component $G_{1}$, which has $n_{1}$ vertices.
So we know it already minimum number edge in a undirected connected graph is $n-1$ . So for Component $G_{1}$ minimum number of edge is $n_{1}-1$.
For component $G_{2}$ ,minimum number edge is $n_{2}-1$.
So on,
For component $G_{k}$ minimum number edge is $n_{k}-1$.
So , Total number of edges in the graph
=$(n_{1}-1)+(n_{2}-1)+(n_{3}-1)+.......+(n_{k}-1)$
=$n_{1}+n_{2}+n_{3}+.......+n_{k}-k$
=$n-k$
For the second case , We get maximum number of edges only when we have one big component with n-k+1 vertices with edges and rest k-1 component with 1 vertices with no edges .
The component with n-k+1 vertices should form a complete graph to make maximum number of edges.
So , Number of edges in a complete graph with (n-k+1) vertices is,
$\binom{n-k+1}{2}$=$\frac{(n-k)(n-k+1)}{2}$
Proof for this can be deduce by contradiction .
Link :- https://math.stackexchange.com/questions/1075692/number-of-edges-in-a-graph-with-n-vertices-and-k-connected-components
