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67 votes

How many graphs on $n$ labeled vertices exist which have at least $\frac{(n^2 - 3n)}{ 2}$ edges ?

- $^{\left(\frac{n^2-n}{2}\right)}C_{\left(\frac{n^2-3n} {2}\right)}$
- $^{{\large\sum\limits_{k=0}^{\left (\frac{n^2-3n}{2} \right )}}.\left(n^2-n\right)}C_k$
- $^{\left(\frac{n^2-n}{2}\right)}C_n$
- $^{{\large\sum\limits_{k=0}^n}.\left(\frac{n^2-n}{2}\right)}C_k$

120 votes

Best answer

Let $a = \frac{n(n-1)}{2}, b = \frac{n^2 -3n}{2}$

Minimum no of edges has to be $\frac{n^2 -3n}{2} = b$.

Maximum no of edges in simple graph = $\frac{n(n-1)}{2} = a$.

So, no of graph with minimum $b$ edges :

$= C(a,b) + C(a,b+1) + C(a,b+2) + \ldots +C(a,a)$

$= C(a,a-b) + C(a,a-(b+1)) + C(a,a-(b+2)) + \ldots +C(a,0)$

$= C(a,n) + C(a,n-1) + C(a,n-2) + \ldots +C(a,0))$$\;\left(\because a-b = n \right)$

$= C\left(\frac{n(n-1)}{2},n\right) + C\left(\frac{n(n-1)}{2}, n-1\right) \\+ C\left(\frac{n(n-1)}{2},n-2\right) + \ldots +C\left(\frac{n(n-1)}{2},0\right)$

$ =\sum\limits_{k=0}^{n} {}^{\left(\frac{n^2-n}{2}\right)}C_k$

Option (**D**).

126 votes

Suppose n=4 labeled vertices.

Max possible edges = $\frac{n(n-1)}{2}$ = $\frac{4*3}{2}$ = 6 edges

At least edges in the graph = $\frac{n(n-3)}{2}$ = $\frac{4*1}{2}$ = at least 2 edges in the graph

Total possible number of graphs = ${}^{6}C_2$ + ${}^{6}C_3$ + ${}^{6}C_4$ +${}^{6}C_5$ +${}^{6}C_6$.

As ${}^{6}C_2$ is same as ${}^{6}C_4$ because both are same as $\frac{6!}{4!2!}$, So ${}^{n}C_r$ = ${}^{n}C_(n-r)$

Re write the the above sequence

${}^{6}C_2$ + ${}^{6}C_3$ + ${}^{6}C_4$ +${}^{6}C_5$ +${}^{6}C_6$ = ${}^{6}C_4$ + ${}^{6}C_3$ + ${}^{6}C_2$ +${}^{6}C_1$ +${}^{6}C_0$ = $$\sum _{k=0}^{n} \frac{n(n-1)}{2}C_k$$

**Hence, option (D) is correct!**

0

3 votes

For n = 1, we need at least -1 edges. This seems troublesome, take some other value.

For n = 2, we need at least -1 edges. Again, take some other value.

For n = 3, we need at least 0 edges.

So, such graphs possible for 3 vertices are:-

$1 + 3 +3+1$ (for 0 edges, for 1 edge, for 2 edges and for 3 edges respectively) => $8$

Look at **Option D**. It is:

${^3C_0} + {^3C_1} + {^3C_2} + {^3C_3}$

=> $1 + 3 +3+1$ => $8$