There are 6 points on a circle. Two triangles | vedclass

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(D) Let the $6$ points on the circle be $P_1, P_2, P_3, P_4, P_5, P_6$. Total number of ways to choose $2$ disjoint sets of $3$ vertices each is given

Vedclass · Accepted Answer

$\frac{3}{10}$

Vedclass · Answer

(D) Let the $6$ points on the circle be $P_1, P_2, P_3, P_4, P_5, P_6$. Total number of ways to choose $2$ disjoint sets of $3$ vertices each is given by $\frac{\binom{6}{3} 	imes \binom{3}{3}}{2!} = \frac{20 	imes 1}{2} = 10$. For the sides of the two triangles not to intersect,the triangles must be formed by choosing $3$ consecutive points for one triangle and the remaining $3$ consecutive points for the other triangle. There are only $2$ such pairs of triangles: $(P_1, P_2, P_3)$ and $(P_4, P_5, P_6)$,or $(P_2, P_3, P_4)$ and $(P_5, P_6, P_1)$,and so on. Specifically,the number of ways to choose $2$ non-intersecting triangles is $2$. Thus,the probability is $\frac{2}{10} = \frac{1}{5}$. Wait,re-evaluating: The total ways to choose $2$ triangles with no common vertices is $\binom{6}{3} = 20$ (choosing $3$ for the first,the rest are fixed). The number of ways to choose $2$ non-intersecting triangles is $2$. Therefore,the probability is $\frac{2}{20} = \frac{1}{10}$.

There are $6$ points on a circle. Two triangles are drawn such that they have no common vertices. What is the probability that none of the sides of the triangles intersect?

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