Please answer in detail________________________________________________
ABC , 6 Years ago
Grade 11
Algebra> Please answer in detail _________________...
1 Answers
Arun
Last Activity: 6 Years ago
i) In total there are 3p points, since each of the 3 lines has 'p' points in it.
ii) To form a triangle we need 3 non collinear points.
So number of triangles that could be formed from 3p points = C(3p, 3)
= (3p)*(3p-1)(3p-2)/3! = p(3p-1)(3p-2)/2
iii) Out of these 3p points, each of the 3 lines contain p points. as such no triangle could be formed from each of these 3 set of 'p' points.
So number of such triangles = 3*C(p,3) = 3*p*(p-1)*(p-2)/3! = p(p-1)(p-2)/2
iv) Thus number of triangles formed = p(3p-1)(3p-2)/2 - p(p-1)(p-2)/2
This simplifies to: 4p³ - 3p²
ii) To form a triangle we need 3 non collinear points.
So number of triangles that could be formed from 3p points = C(3p, 3)
= (3p)*(3p-1)(3p-2)/3! = p(3p-1)(3p-2)/2
iii) Out of these 3p points, each of the 3 lines contain p points. as such no triangle could be formed from each of these 3 set of 'p' points.
So number of such triangles = 3*C(p,3) = 3*p*(p-1)*(p-2)/3! = p(p-1)(p-2)/2
iv) Thus number of triangles formed = p(3p-1)(3p-2)/2 - p(p-1)(p-2)/2
This simplifies to: 4p³ - 3p²
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