# what is solution of equation arrangements(x+1,2)=combinations(x-1,2)+12?

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A(x+1,2) = C(x-1,2)+12

We'll recall the formula of arrangements:

A(x+1,2) = (x+1)!/(x+1-2)! = (x+1)!/(x-1)!

But (x+1)! = (x-1)!*x*(x+1) => (x+1)!/(x-1)! = (x-1)!*x*(x+1)/(x-1)!

We'll reduce like terms:

A(x+1,2) = x*(x+1)

We'll recall the formula of combinations:

C(x-1,2) = (x-1)!/2!*(x-1-2)! =(x-1)!/2!*(x-3)!

But (x-1)! = (x-3)!*(x-2)*(x-1) =>C(x-1,2) = (x-3)!*(x-2)*(x-1)/2!*(x-3)!

But 2! = 1*2 = 2

We'll reduce like terms:

C(x-1,2) = (x-2)*(x-1)/2

We'll re-write the equation:

A(x+1,2) = C(x-1,2)+12 <=> x*(x+1)=(x-2)*(x-1)/2 + 12

We'll remove the brackets and we'll multiply by 2 both sides:

2x^2 + 2x = x^2 - 3x + 2 + 24

We'll move all terms to the left side:

x^2 + 5x - 26 = 0

We'll apply quadratic formula:

x1 = (-5+sqrt(25+104))/2

Since sqrt129 is an irrational number, then the solutions of the equation are irrational, which is impossible because x must be a natural number.

**Therefore, the given equation has no natural solutions.**