Question

Consider a 20-sided convex polygon K, with vertices A1,A2, . . . ,A20 in that order. Find the number of ways in which three sides of K can be chosen so that every pair among them has at least two sides of K between them. (For example (A1A2,A4A5,A11A12) is an admissible triple while (A1A2,A4A5,A19A20) is not.)

Answer 1

method 1 :- this method like as trick .
For a n - sided polygon , the number of such triplet is n(n -7)(n -8)/6 where n≥ 9
so, here, n = 20
number of ways = 20 × (20-7) × (20 - 8)/6
= 20 × 13 × 12/6 = 520

method 2 :- Here we use concept of permutation and combination for solving this question .

solution :- we have a convex polygon K, with vertices A1 , A2 , A3 .......A20 .

Total number of ways to select three sides in 20 sides of K polygon = 20C3
= 20!/17!3! = 20× 19×18/6 = 1140

now, no of ways of selection when exactly three sides , are common ( I mean {A1A2 , A2A3 ,A3A4} type all selection ) = 20

now , number of ways of selection when exactly two sides are common { i mean if we selected {A1A2, A2A3} then we can't select A3A4 and A1A20 } = 20 × 16 = 320

again, number of ways of selection when the selected two sides are A1A2 & A3A4 ( I mean one side gap between them ) = 20

Two cases possible for 3rd side.
case1 :- 3rd side selected at a gap of two from above selected two sides.
{ I mean if we take A1A2 , A3A4 , then we can't choose 3rd side A3A4, A4A5 , A20A1 , A19A20 }
number of ways of 3rd side section = 20 -6 -1 = 13
required ways = 20 × 13 = 260

case2 :- 3rd sides is at a gap of one from any selected side = 20

so, number of ways = 1140 - 20 - 320 - 260 - 20 = 520