Combinations: How many handshakes? I need help with the following question which I cannot seem to solve:

17 students are sitting in a circle. Each person shakes hands with everyone but his/her neighbours. How many handshakes have been exchanged?

My approach: no. of ways $ = 1 + 2 + 3 + ... + 14 = 7(15) = 105 $.

Apparently my answer is wrong (correct ans is 119). But I can't seem to understand why. Could someone please explain?
Each of 17 people shakes hands with 14 people (all except themselves and their 2 neighbors), so there are

$$\frac{17\times 14}{2} = 119$$

handshakes (dividing by 2 to account for symmetry, as you would otherwise count "$A$ shaking hands with $B$" and "$B$ shaking hands with $A$" as distinct events).
I will explain for 6, same logic applies for all the numbers. 

first, 
if 6 persons shakes will all in the party, then number of shakes will be: 

A  B  C  D  E  F

A can shake with only rest of 5 persons(B,C,D,E,F) i.e, right side ppl. = 5

B can shake with only his right sides(C,D,E,F) ppl(not with A bcos,A already shook with B just now. so no repeat) = 4 

C can shake with only his right sides(D,E,F) ppl(not with A&amp;B bcos,A&amp;B already shook with C just now. so no repeat) = 3

D can shake with only his right sides(E,F) ppl(not with left side ppl  bcos, they already shook with D just now. so no repeat) = 2 

E can shake with only his right sides(F) ppl(not with left side ppl  bcos, they already shook with E just now. so no repeat) = 1

as all of the persons shook hand with F, He dont shake with any one now. 

So total possible shakes for 6 persons are = 5+4+3+2+1 => 15

now in questions it is given that, No one shakes hand with neighbors. 

Imagine ppls sitting around table

        A   B
      F       C
        E   D   


Please treat above image as 6 Ppl around circle :
So now, 

Invalid Shakes:

A cannot shake with B,F . I am representing it as below :
  A  -no- B &amp; F    => 2 invalid shakes.
  B  -no- C only. Bcos  B -no- A is covered as part of first case.
                   =>  1 invalid shake
  C  -no- D only.  =>  1 invalid shake
  D  -no- E only.  =>  1 invalid shake
  E  -no- F only.  =>  1 invalid shake
  F dont shake with A is already covered as part of first invalid shakes.    

so total invalid shakes are 2+1+1+1+1 = 6   

so for 6 ppl:  out of  total 15 , 6 are invalid. So 15-6 = 9  

on the same line :   

for 17 ppl:  out of  total (16+15+...+1)=136 , 17 are invalid. So 136-17 = 119

shortcut :   For N ppl, Total shakes are  (N-1)(first term + last term)/2  Arithmetic progression sum. out of which Invalids are N.
So valid shakes =>  (Total - N)