What is the natural number n if C(2n-3,2)=3?

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neela | High School Teacher | (Level 3) Valedictorian

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What is the natural number n if C(2n-3,2) = 3.

We know that C(2n-3 , 2) = (2n-3)!/{(2n-3 - 2)!*2!} = 3C(2n-3) = (2n-3)!/(2n-5)!*2! = 3

C(2n-3 , 2) = (2n-3)(2n-4)(2n-5)!/(2n-5)!*2 = 3

=> (2n-3)(2n-4)/2 = 3

(2n-3)(2n-4) = 3*2 .

4n^2 -14n+12 = 6.

4n^2-14n +12-6 = 0.

4n^2-14n+6 = 0

2n^2-7n +3 = 0.

(2n-1)(n-3) = 0.

Therefore  n = 3.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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By definition, C(2n-3,2) = (2n-3)!/2!*(2n-3-2)!

C(2n-3,2) = (2n-3)!/2!*(2n-5)!

But (2n-5)! = (2n - 5)!(2n - 4)(2n - 3)

2! = 1*2 = 2

C(2n-3,2) = (2n - 5)!(2n - 4)(2n - 3)/2*(2n - 5)!

We'll simplify and we'll get:

C(2n-3,2) = (2n - 4)(2n - 3)/2

Now, we'll solve the equation:

(2n - 4)(2n - 3)/2 = 3

(2n - 4)(2n - 3) = 6

We'll remove the brackets:

4n^2 - 6n - 8n + 12 = 6

4n^2 - 14n + 6 = 0

We'll divide by 2:

2n^2 - 7n + 3 = 0

We'll apply the quadratic formula:

n1 = [7+sqrt(49 - 24)]/4

n1 = (7 + 5)/4

n1 = 3

n2 = (7-5)/4

n2 = 1/2

Since n2 is not a natural number, we'll accept as solution just n = 3.

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