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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 -

455

Option 2 -

443

Option 3 -

12

Option 4 -

419

0 5 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    a month 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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