Question

solve this equation 2nC2=15 (its a combination)

Answer 1

Answer

The value of n is 3

Given

$^{2n} C _{2} = 15$

To Find

• The value of n

Solution

$^{2n} C_{2} = 15 \\ \implies \ \frac{(2n)!}{(2! )(2n - 2)! } = 15 \\ \implies \frac{2n \times (2n - 1) \times (2n - 2)! }{2 \times 1 \times (2n - 2)! } = 15 \\ \implies n(2n - 1) = 15 \\ \implies 2{n}^{2} - n - 15 = 0$

Now Solving the quadratic equation we will have the value of n

$\implies2 {n}^{2} - n - 15 = 0 \\ \implies2 {n}^{2} - 6n + 5n - 15 = 0 \\ \implies2n(n - 3) + 5(n - 3) = 0 \\ \implies(n - 3)(2n + 5) = 0$

Now we have

$n - 3 = 0 \\ \implies n = 3$

And

$2n + 5 = 0 \\ \implies n = \frac{ - 5}{2}$

Since the value of n can never be negative or fraction so n ≠ -5/2

Thus the value of n is 3

Answer 2

Solutions:

²nC2 = 15

(2n)/ (2n)(2n - 2)

2n × (2n - 1) × (2n +2) / 2 × 1 × (2n - 2) = 15

n(2n - 2) = 15

2n² - n - 15 = 0

2n² - 6n + 5n - 15 = 0

2n(n - 3) + 5(n - 3) = 0

(2n + 5) (n - 3) = 0

n - 3 = 0

n = 3

2n + 5 = 0

n = -5/2.

silentlover45.❤️