let n(>9) be the number of vertices and Y is the number of connectives.connectives which are connected from one vertex to other vertices(connectives=no two lines coincide with each other) then y is divisible by
option
a) n
b) n-1
c) n-2
d) none of these

• in any polygon no. of of diagonal will be nc2-n. bcoz total no of edges will be nc2 which contain edges and diagonal so we have to eliminate no of side which will be equal to nc2-n which is equal to n*((n-1)/2)-n which is not divisible by any of these option so none of these...........check for n>9
• 10 years agoHelpfull: Yes(5) No(2)
• If n denotes the No of vertices of polygon then the ans will be No of diagonal divisible by i.e n*(n-3)/2
  hence ans a) n
• 10 years agoHelpfull: Yes(2) No(5)
• See, lets move step by step...
  For n=10, y= 10C2-10 = 35
  For n=11, y= 11C2-11 = 44
  For n=12, y= 12C2-12 = 54
  For n=13, y= 13C2-13 = 65
  For n=14, y= 14C2-14 = 77
  For n=15, y= 15C2-15 = 90
  and so on..
  So what we are seeing here is for all odd values of n, y is divisible by n. Then what about the even values of n.?
  If we closely observe the results, we'll for all even values of n. y is actually divisible by n-3..
  As,
  In case of n=10, y=35 is divisible by 10-3=7,
  In case of n=12, y=54 is divisible by 12-3=9,
  In case of n=11, y=77 is divisible by 14-3=11
  That is we don't have a generalize formula for both even and odd values of n, but a generalize formula for individual values of even or odd n.
  Hence, I guess option D as correct..:)
  
  
• 10 years agoHelpfull: Yes(2) No(0)
• friends n*((n-1)/2))-n can be written as n*(n-3)/2 so its divisible by n
• 10 years agoHelpfull: Yes(0) No(2)
• for n=14 is it divisible by either n,n-1,.....please can u check......answer is D.none of these........
• 10 years agoHelpfull: Yes(0) No(0)
• If n denotes the No of vertices of polygon then the ans will be No of diagonal divisible by i.e n*(n-3)/2
  hence ans a) n
  
• 9 years agoHelpfull: Yes(0) No(0)