In how many ways 3 vertices can be selected from a polygon of 10 sides so that no vertices are consequentAns: 50

the ques means that no two things r together. let we do it for n sides.

so sol is   nC₃ - ( 3 things together)  - (2 things together and 2rd away)

let choose any set, say    (1,2,*)      *= denotes any other no.

as 1 n 2 r together so it these types r to be eliminated.

now (1 ,2 ,*)  * can be filled by (n-2) ways or nos. see that it also includes case when all nos r consequent( 3 together)

there r 10 sets like this with (1,2,*)   (2,3,*)...............(n,1,*)

so total of these types is n*(n-2)

see it includes 3 togher cases two times     eg.    in both set (1,2,*) and (2,3,*)   set  (1,2,3) is formed.

it doesnot matter that order is diff.

so they r subtracted twice so again 1 time they should be added to solution.

now, no of 3 togher,

1,2,3     2,3,4    3,4,5  ...............  n,1,2       total n ways

so final solution is

nC₃ - n(n-2)  + n          ans.

putting n=10

10C₃- 10*8 + 10 =50

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