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Bunuel
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The solution should be Option D.

Can be solved using permutations and combinations formula to choose 2 more vertices from the remaining 4 points.

Can be written as

4C2 = 4!/2!(4-2)!
=4!/2!*2!
which comes to 6.
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Since one of the point P is fixed we need 2 points from the remaining 4 points

So 4C2 = 6

Hence answer = D
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The total triangles possible : 5C3
number of triangles with P as vertex = (total triangle possible) - (No. of triangle without P as vertex)
= 5C3 - 4C3
= 10 - 4
= 6
Option D is answer.
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