Actually, the question should be: simple graphs are possible on six labeled vertices in which the number of edges is 'odd'.
Answer: we can do with recurrence method
for n=3 , no. of simple graphs are possible = 2^(3C2)= 2^3 =8 // [maximum edges with n vertices in simple graph: n(n-1)/2= 3*2/2 = 3]
Now, as @Mk Utkarsh solved, we can calculate no. of simple graphs with odd edges like:
C(3,1)+C(3,3)= 3+1= 4 // [this is combination form] // 2^2 ( 2^3 /2 )
for n=4 , no. of simple graphs are possible = 2^(4C2)= 2^6 =64
Now, we can calculate no. of simple graphs with odd edges( Total edges =6) like:
C(6,1)+C(6,3)+C(6,5) = 6+20+6= 32 // ( 2^6 / 2 = 2^6-1 = 2^5 =32)
for n=5 , no. of simple graphs are possible = 2^(5C2)= 2^10 =1024
Now, we can calculate no. of simple graphs with odd edges( Total edges =10) like:
C(10,1)+C(10,3)+C(10,5)+C(10,7)+C(10,9) = 10+120+252+120+10= 512 // (2^10-1 = 2^9 =512)
NOW, YOU CAN SEE THAT SIMPLE GRAPHS WITH ODD NO. OF EDGES OVER 'n' VERTICES ARE:
=[ 2^(nC2) ] / 2 .
Now put n=6
no. of simple graphs with odd no. of edges over 6 vertices = [2^(6C2)] /2
= (2^15) / 2
= 2^14 graphs