Question

There are n Distinct points on the circumference of a circle.The Number of pentagons that can be formed with these points as vertuces is equal to the number of possible triangles.Then Value of n is?​

Answer 1

$\huge{\underline{\sf{\red{Answer}}}}$

Given:

Number of possible pentagons = Number of possible triangles

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$\huge{\underline{\underline{\green{\tt{Formula}}}}}$

$\star$ Number of pentagons formed by joining n distinct points on the circumference of the circle = $^n C_5$

$\star$ Number of Triangles formed by joining n distinct points on the circumference of the circle = $^n C_3$

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SoluTion:

$\huge{^n C_5 = ^n C_3 }$

$\leadsto\dfrac{\cancel{n!}}{(n-5)! \: 5!} = \dfrac{\cancel{n!}}{(n-3)! 3!}$

$\leadsto{(n-3)! 3! = (n-5)!5!}\\ \\ {\leadsto{(n-3)(n-4)\cancel{(n-5)(n-6)(n-7)..... }3! = (n-5)\cancel{(n-6)(n-7).........} 5!}}\\ \\ {\leadsto{(n^2 -7n+12)3\times 2\times = 5 \times 4\times 3\times 2}} \\ \\ {\leadsto{n^2-7n+12-20=0}}\\ \\ {\leadsto{n^2-7n-8}}$

By solving the Quadratic

We get n= 8,-1

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Answer 2

$\mathfrak{\huge{\orange{\underline{\underline{Answer :}}}}}$

Given:

Number of possible pentagons = Number of possible triangles

___________________________________

$\huge{\underline{\underline{\green{\tt{Formula}}}}}$

$\star$ Number of pentagons formed by joining n distinct points on the circumference of the circle = $^n C_5$

$\star$ Number of Triangles formed by joining n distinct points on the circumference of the circle = $^n C_3$

___________________________

SoluTion:

$\huge{^n C_5 = ^n C_3 }$

$\leadsto\dfrac{\cancel{n!}}{(n-5)! \: 5!} = \dfrac{\cancel{n!}}{(n-3)! 3!}$

$\leadsto{(n-3)! 3! = (n-5)!5!}\\ \\ {\leadsto{(n-3)(n-4)\cancel{(n-5)(n-6)(n-7)..... }3! = (n-5)\cancel{(n-6)(n-7).........} 5!}}\\ \\ {\leadsto{(n^2 -7n+12)3\times 2\times = 5 \times 4\times 3\times 2}} \\ \\ {\leadsto{n^2-7n+12-20=0}}\\ \\ {\leadsto{n^2-7n-8}}$

By solving the Quadratic

We get n= 8,-1

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