Let P1, P2 ...., P15 be 15 points on a circle. The number of distinct triangles formed by points Pi, Pj, pk such that i + j + k 15, is

Option 1 - <p>455</p>
Option 2 - <p>443</p>
Option 3 - <p>12</p>
Option 4 - <p>419</p>
7 Views|Posted 6 months ago
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1 Answer
A
6 months ago
Correct Option - 2
Detailed Solution:

Required number =  - number of solution of x1 + x2 + x3 = 15, where x1 < x2 < x3

For x2 = x1 + a, a   1

->3x1 + 2a + b = 15

Coefficient of x15 in

( x 3 + x 6 + x 9 + x 1 2 + x 1 5 )

( x 1 + x 2 + x 3 + . . . . . + x 1 0 ) = 1 2

Required number = 455 – 12 = 443

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