## 3 Answers

$N$ vertices and $K$ components.

(Component means they are non connected subgraphs)

We want maximum edges in total graph. How to get maximum edges ?

**To get maximum, take one vertex each for each component, except last component.**

Now $k-1$ components have $1$ vertex each and so no edges.

The last component has $n-(k-1)$ vertices.

So make the last component complete.

i.e., It has ${}^{n-(k-1)} C_ 2 =\frac{ (n-k) (n-k+1) }{ 2}$** ** edges.** **

*Must do a similar model qsn on forest: https://gateoverflow.in/580/gate1992_03-iii*

### 4 Comments

Hopefully it should be clear that in any such graph all components will be complete, i.e., have all possible edges. Thus the only remaining question is how large each component should be?

If there are two components with $a$ and $b$ vertices, $a>1, b> 1$, then together they can have at most

$\binom{a}{2} + \binom{b}{2} = \frac{1}{2} \left(a^2 - a + b^2 - b \right)$ edges.

However, if we place all but one of the vertices in a single component, we could have

$\binom{a+b-1}{2} + \binom{1}{2} = \frac{1}{2} \left(a+b-1\right)\left(a + b - 2\right)$

$=\frac{1}{2} \left(a^2 + 2ab -3a + b^2 - 3b + 2 \right)$ edges.

Subtracting the first quantity from the second gives

$\frac{1}{2}\left(\left(2ab - 3a - 3b +2\right) - \left(-a - b \right)\right) \\=ab-a-b+a \\= (a-1)(b-1) \text{ which is } > 0$

Hence it is better not to have two components with multiple vertices.

This leaves us with the answer that all components should have one vertex except one, which will have $n-k+1$ vertices, for a total of $\binom{n-k+1}{2}$ edges.

In simple connected graph, with number of edges as $e,$ we have

$(n-1) \leq e \leq n. \frac{(n-1)}{2}$

In simple disconnected graph with $k$ components and number of edges as $e,$ we have

$(n-k) \leq e \leq (n-k).\frac{(n-k+1)}{2 }$

Note: Put $k=1$ then it will be connected graph .

Reference @ http://www.quora.com/What-is-the-maximum-number-of-edges-in-graph-with-n-vertices-and-k-components

Another read @ http://stackoverflow.com/questions/24003861/maximum-number-of-edges-in-undirected-graph-with-n-vertices-with-k-connected-com

### 9 Comments

if u want maximum number of nodes and k vertices then u should have k-1 components. with only one vertices and only one should contain remaining vertex ,

like

if i take 6 vertex and have to make 4 component then i will make the 4 component in this way ,

1 vertex 1 vertex 1 vertex 3 vertex.

and now i will make the last one complete graph . then it will have maximum number of edges.

so if u have n vertices and k component then just give (k-1) component one vertex and the remaining will be (n-(k-1)) now make that bigger one a complete graph . i.e ( n-k+1) ( n-k-1+1)/2 complete graph formula . n(n-1)/2 = (n-k)(n-k+1)/2