VCP equation

Solve the following equation with variations, combinations and permutations:

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

Result

x =  3

Solution:

4 V(2,x)3 (2x+1)xP(2)=0  4 x (x1)3 (x+1) x/2x 2=0 2.5x27.5x=0  a=2.5;b=7.5;c=0 D=b24ac=7.5242.50=56.25 D>0  x1,2=b±D2a=7.5±56.255 x1,2=1.5±1.5 x1=3 x2=0   Factored form of the equation:  2.5(x3)x=0x>0 x=x1=3 C2(4)=(42)=4!2!(42)!=4321=6  b=(2x+1)=(23+1)=6 y=4 3 2 1/(1)3 bx 2 1=4 3 2 1/13 63 2 1=04 \ V(2,x)-3 \ { { 2 } \choose x+ 1 } - x P(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 = 0x>0 \ \\ x = x_{ 1 } = 3 \ \\ C_{{ 2}}(4) = \dbinom{ 4}{ 2} = \dfrac{ 4! }{ 2!(4-2)!} = \dfrac{ 4 \cdot 3 } { 2 \cdot 1 } = 6 \ \\ \ \\ 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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