find the number of triangles that can be formed using 14 points in aplane such that 4 points are colliner​

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1. [tex]\large\underline{\sf{Solution-}}[/tex]

   We know,

   ★ If there are n non collinear points in the plane, then number of triangles formed is given by

   [tex]\rm :\longmapsto\:Number \: of \: triangles = \: ^nC_3[/tex]

   and we know,

   [tex]\rm :\longmapsto\:^nC_r \: = \: \dfrac{n!}{r!(n-r)!} [/tex]

   Now,

   ★ It is given that, 14 points in a plane out of which 4 points are collinear.

   ★ Number of triangles formed by joining 14 points taken 3 at a time is

   [tex]\rm :\longmapsto\:Number \: of \: triangles = \: ^{14}C_3[/tex]

   [tex] \rm \: \: = \: \dfrac{14!}{3!(14 – 3)!} [/tex]

   [tex] \rm \: \: = \: \dfrac{14!}{3! \: \: 11!} [/tex]

   [tex] \rm \: \: = \: \dfrac{14 \times 13 \times 12 \times 11!}{3 \times 2 \times 1\: \: 11!} [/tex]

   [tex] \rm \: \: = \: 28 \times 13[/tex]

   [tex] \rm \: \: = \: 364[/tex]

   Now,

   ★ Number of triangles formed by joining 4 points taken 3 at a time is

   [tex]\rm :\longmapsto\:Number \: of \: triangles = \: ^{4}C_3[/tex]

   [tex] \rm \: \: = \: \dfrac{4!}{3!(4 – 3)!} [/tex]

   [tex] \rm \: \: = \: \dfrac{4 \times 3!}{3! \: \: 1!} [/tex]

   [tex] \rm \: \: = \: 4[/tex]

   ★ But 4 points are collinear which don’t form a triangle when taken 3 at a time.

   So,

   ★ Required number of triangles = 364 – 4 = 360.

   _____________________________________

   Additional Information :-

   [tex]\rm :\longmapsto\:^nC_r \: = \: \dfrac{n}{r} \: ^{n – 1}C_{r – 1}[/tex]

   [tex]\rm :\longmapsto\:^nC_0 = ^nC_n = 1[/tex]

   [tex]\rm :\longmapsto\:^nC_1 = ^nC_{n – 1} = n[/tex]

   [tex]\rm :\longmapsto\:^nC_r \: + \: ^nC_{r – 1} \: = \: ^{n + 1}C_r[/tex]

   [tex]\rm :\longmapsto\:\dfrac{^nC_r}{^nC_{r – 1}} = \dfrac{n – r + 1}{r} [/tex]

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