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

P(x , 2) = x! / (x - 2)!

Now C(x,2) + P(x,2) = 30

=> x!/ ( 2! *(x - 2)!) + x! / (x - 2)! = 30

=> x*(x - 1)/2! + x*(x - 1) = 30

=> x*(x - 1)/2 + x*(x - 1) = 30

=> x*(x - 1) + 2*x*(x - 1) = 60

=> x^2 - x + 2x^2 - 2x = 60

=> 3x^2 - 3x = 60

=> x^2 - x = 20

=> x^2 - 5x + 4x - 20 = 0

=> x(x - 5) + 4(x - 5) = 0

=> (x + 4)(x - 5) =0

x can be -4 or 5.

A negative value of x has no meaning , therefore it is ignored. So x = 5.

**The required value of x = 5.**

check:

C( 5, 2) + P(5 , 2) = 30.

We assume that C stands for combination and A stands for arrangements.

The number of ways selecting 2 things fro x things is x(x-2)/2.

The number of arrangements arrangement of x things in two places is x(x-2).

Therefore C(x,2)+ A(x,2) = 30 => x(x-1)/2 + x(x-1) = 30

x(x-1)+2x(x-1) = 60.

x^2-x +2x^2-2x = 60.

3x^2 -3x -60 = 0.

We divide by 3.

x^2-x-20 = 0.

(x-5)(x+4) = 0.

x-5 = 0, or x+4 = 0.

So x= 5.

We'll use combinatorics and we'll write:

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

But x! = (x-2)!*(x-1)*x

2! = 1*2 = 2

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

We'll simplify and we'll get:

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

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

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

We'll simplify and we'll get:

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

We'll re-write the given equation:

(x-1)*x/2 + (x-1)*x = 30

We'll factorize by (x-1)*x:

(x-1)*x(1/2 + 1) = 30

3(x-1)*x/2 = 30

We'll divide by 3:

(x-1)*x/2 = 10

We'll multiply by 2:

(x-1)*x = 20

We'll remove the brackets and we'll subtract 20:

x^2 - x - 20 = 0

We'll apply quadratic formula:

x1 = [1 + sqrt(1+80)]/2

x1 = (1 + 9)/2

x1 = 5

x2 = (1-9)/2

x2 = -4

**Since x has to be natural, we'll reject the second solution, so the solution of the equation is x = 5.**