# solve eq 4(x+1)C4=15(x+1)A2

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You should re-write the members of equation using the following factorial formulas, such that:

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

`A_n^k = (n!)/((n - k)!)`

Reasoning by analogy, yields:

`C_(x+1)^4 = ((x+1)!)/(4!(x + 1 - 4)!) => C_(x+1)^4 = ((x+1)!)/(4!(x - 3)!)`

`A_(x+1)^2 = ((x+1)!)/((x - 1)!)`

Replacing `((x+1)!)/(4!(x - 3)!)` for `C_(x+1)^4` and `((x+1)!)/((x - 1)!)` for `A_(x+1)^2` yields:

`4*((x+1)!)/(4!(x - 3)!) = 15*((x+1)!)/((x - 1)!)`

`4*((x+1)!)/(1*2*3*4(x - 3)!) = 15*((x+1)!)/((x - 3)!(x - 2)(x - 1))`

Reducing duplicate factors both sides yields:

`1/6 = 15/((x - 2)(x - 1)) => 6*15 = (x - 2)(x - 1)`

Performing the multiplication to the right side, yields:

`x^2 - 3x + 2 - 90 = 0`

`x^2 - 3x - 88 = 0 => x_(1,2) = (3+-sqrt(9 + 352))/2`

`x_(1,2) = (3+-19)/2 => x_1 = (3 + 19)/2 => x_1 = 11`

`x_2 = (3 - 19)/2 => x_2 = -8`

Since x needs to be a natural number, you cannot validate the value `x = -8.`

**Hence, evaluating the solution to the given equation, yields x = 11**.