You need to remember the factorial formula for combinations such that:

C_n^k = (n!)/(k!*(n-k)!)

Hence, C_(n+2)^4 = ((n+2)!)/(4!*(n+2-4)!)

C_(n+2)^4 = ((n+2)!)/(4!*(n-2)!)

C_n^2 = (n!)/(2!*(n-2)!)

Hence, writing the equation in factorial form yields:

((n+2)!)/(4!*(n-2)!) = 6(n!)/(2!*(n-2)!)

Reducing by (2!*(n-2)!) yields:

((n+2)!)/(3*4) = 6(n!) => (n+2)! = 12*6*n!

You may write (n+2)! in terms of n! such that:

(n+2)! = n!*(n+1)(n+2)

Writing the equation using (n+2)! = n!*(n+1)(n+2) yields:

`n!*(n+1)(n+2) = 12*6*n!`

`` Reducing by n! yields:

`(n+1)(n+2) = 72`

You need to open the brackets to the left side such that:

`n^2 + 2n + n + 2 = 72`

Subtracting 72 both sides yields:

`n^2 + 3n - 70 = 0`

You need to remember quadratic formula that helps finding the zeroes of quadratic equations.

`n_(1,2) = (-3+-sqrt(9 + 280))/2 =gt n_(1,2) = (-3+-sqrt(289))/2`

`n_(1,2) = (-3+-17)/2 =gt n_1 = (-3+17)/2 = 7`

`n_2 = (-3-17)/2 = -10`

**Since n need to be positive, hence the only solution to the equation is n = 7.**

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