# science of combinatorics(x+2)c2=14

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You need to solve for x the equation `(x+2)C2 = 14` , hence, you should remember factorial formula such that:

`nCk = (n!)/(k!*(n-k)!)`

`(x+2)C2 = ((x+2)!)/(2!*(x+2-2)!)`

`(x+2)C2 = ((x+2)!)/(2!*(x)!)`

You may write `((x+2)!) = x!*(x+1)(x+2)` such that:

`(x+2)C2 = (x!*(x+1)(x+2))/(2!*(x)!)`

You need to reduce like terms such that:

`(x+2)C2 = ((x+1)(x+2))/(1*2)`

You may substitute `((x+1)(x+2))/(1*2)` for `(x+2)C2` in equation such that:

`((x+1)(x+2))/(1*2) = 14`

`x^2 + 2x + x + 2 = 28`

`x^2 + 3x - 26 = 0`

`x_(1,2) = (-3+-sqrt(9+104))/2`

`x_(1,2) = (-3+-sqrt113)/2`

**Since `x_1` and `x_2` are not natural numbers, hence, there is no solution to equation `(x+2)C2 = 14.` **

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