there is a bigmistake in the problem. you didn't mention if the things were distinct or not. ;)
assuiming they are distinct (cause it is the only case the problem cn be solved given the conditions).
we have A = (x+2)!.
B = (x!)(x-11)!
C = (x-11)!
using the relation given we get -
(x+2)! = 182 (x!).
or, (x+1)(x+2) = 182.
or, x^2 + 3x + 2 = 182.
or x^2 + 3x - 180 = 0.
x^2 + 15x - 12x - 180 = 0.
(x+15)(x-12) = 0.
since x is a natural number, x = 12.