Home › Forums › Math Olympiad, I.S.I., C.M.I. Entrance › Combinatorics › Combinatorics

- This topic has 1 reply, 2 voices, and was last updated 2 weeks, 4 days ago by Shirsendu Roy.

- AuthorPosts
- June 22, 2020 at 1:14 pm #74593Khushi SharmaParticipant
<p style=”text-align: left;”>Consider a 20-sided convex polygon K, with vertices 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 is an admissible triple while is not.)</p>

June 27, 2020 at 11:39 pm #75057Shirsendu RoyModeratorFor convex polygon K, with vertices A1, A2, A3,….,A20.

Total number of ways to select three sides in 20 sides of k polygon ={20 \choose 3}

=\frac{20!}{17!3!}=\frac{20\times19\times18}{6}=1140

now, number of ways of selection when exactly three sides, are common ({A1A2,A2A3,A3A4} type all selection)=20

now, number of ways of selection when exactly two sides are common{A1A2,A2A3} then we can’t select A3A4 and A1A20}=20 \times 16=320

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

Two cases possible for third side

case1 third side selected at a gap of two from above selected two sides

{if we take A1A2,A3A4 then we can not choose 3rd side A3A4, A4A5,A20A1, A19A20}

number of ways of third side section=20-6-1=13

required ways=20 \times 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

- AuthorPosts

- You must be logged in to reply to this topic.