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

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Permutations and Combinations

Let

P_{1}, P_{2} \dots, 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:

Answer: (B)

The total number of triangles is

(3 15) = 455

. Those excluded satisfy

i + j + k = 15

with

1 \leq i < j < k \leq 15

. Counting triples of distinct values summing to

15

: with smallest part

1

there are

5

, with smallest part

2

there are

4

, with smallest part

3

there are

2

, and with smallest part

4

there is

1

, giving

12

such triangles. Hence the required count is

455 - 12 = 443

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option