# VCP equation

Solve the following equation with variations, combinations and permutations:

4 V(2,x)-3 C(2,x+ 1) - x P(2) = 0

Correct result:

x =  3

#### Solution:

$C_{{ 2}}(4) = \dbinom{ 4}{ 2} = \dfrac{ 4! }{ 2!(4-2)!} = \dfrac{ 4 \cdot 3 } { 2 \cdot 1 } = 6 \ \\ 4 \ V(2,x)-3 \ { { 2 } \choose x+ 1 } - x P(2)=0 \ \\ 4 * x*(x-1)-3 * (x+1)*x/2- x*2=0 \ \\ \ \\ 4 \cdot \ x \cdot \ (x-1)-3 \cdot \ (x+1) \cdot \ x/2- x \cdot \ 2=0 \ \\ 2.5x^2 -7.5x=0 \ \\ \ \\ a=2.5; b=-7.5; c=0 \ \\ D=b^2 - 4ac=7.5^2 - 4\cdot 2.5 \cdot 0=56.25 \ \\ D>0 \ \\ \ \\ x_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ 7.5 \pm \sqrt{ 56.25 } }{ 5 } \ \\ x_{1,2}=1.5 \pm 1.5 \ \\ x_{1}=3 \ \\ x_{2}=0 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 2.5 (x -3) x=0 \ \\ x>0 \ \\ x=x_{1}=3 \ \\ b={ { 2 } \choose x+1 }={ { 2 } \choose 3+1 }=6 \ \\ y=4 \cdot \ 3 \cdot \ 2 \cdot \ 1/(1)-3 \cdot \ b-x \cdot \ 2 \cdot \ 1=4 \cdot \ 3 \cdot \ 2 \cdot \ 1/1-3 \cdot \ 6-3 \cdot \ 2 \cdot \ 1=0$

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