First slide

Combinations

Question

There are three coplanar parallel lines. If any p points are taken on each the lines, the maximum number of triangles with vertices at these points is

Very difficult

A

$3 p^{2} (p - 1) + 1$
B

$3 p^{2} (p - 1)$
C

$p^{2} (4 p - 3)$
D

None of these

Solution

The number of triangles with vertices on different lines
$=^{p} C_{1} \times^{p} C_{1} \times^{p} C_{1} = p^{3}$ .

The number of triangles with 2 vertices on one line and the third vertex on any of the other two lines $=^{3} C_{1} {^{p} C_{2} \times^{2 p} C_{1}} = 6 p \frac{p (p - 1)}{2}$

$∴$ The required number of triangles $= p^{3} + 3 p^{2} (p - 1)$ .

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