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