Let P_{1}, P_{2}, ldots, P_{15} be 15 points | vedclass

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(C) Total number of triangles formed by $15$ points is ${}^{15}C_{3} = \frac{15 	imes 14 	imes 13}{3 	imes 2 	imes 1} = 455$.We need to find the n

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$443$

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(C) Total number of triangles formed by $15$ points is ${}^{15}C_{3} = \frac{15 	imes 14 	imes 13}{3 	imes 2 	imes 1} = 455$.We need to find the number of triangles such that $i+j+k 
eq 15$,where $1 \leq i First,we count the number of sets $(i, j, k)$ such that $i+j+k = 15$ with $1 \leq i - If $i=1$: $j+k=14$. Possible $(j, k)$ are $(2, 12), (3, 11), (4, 10), (5, 9), (6, 8)$. ($5$ cases)- If $i=2$: $j+k=13$. Possible $(j, k)$ are $(3, 10), (4, 9), (5, 8), (6, 7)$. ($4$ cases)- If $i=3$: $j+k=12$. Possible $(j, k)$ are $(4, 8), (5, 7)$. ($2$ cases)- If $i=4$: $j+k=11$. Possible $(j, k)$ is $(5, 6)$. ($1$ case)Total cases where $i+j+k=15$ is $5+4+2+1 = 12$.Thus,the number of triangles such that $i+j+k 
eq 15$ is $455 - 12 = 443$.

Let $P_{1}, P_{2}, \ldots, P_{15}$ be $15$ points on a circle. The number of distinct triangles formed by points $P_{i}, P_{j}, P_{k}$ such that $i+j+k \neq 15$ is:

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