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.

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