Coefficients of \((2 r+4)\) th term and \((r-2)\) th term are equal in expansion of \((1+x)^{18}\) then \(r=\) (a) 4 (b) 5 (c) 6 (d) 7

Using binomial theorem, the general term of expansion of \((a+b)^n\) can be written as: \[_k = \binom{n}{k-1} a^{n-(k-1)} b^{k-1} \] In the given problem, the expansion of \((1+x)^{18}\) can be written in general as: \[_k = \binom{18}{k-1} 1^{18-(k-1)}x^{k-1} \] Now, given that the coefficients of the \((2 r+4)\)th term and the \((r-2)\)th term are equal. So, \[\binom{18}{(2r+4)-1} = \binom{18}{(r-2)-1} \]

Simplify the equation and solve for the value of r: \[\binom{18}{2r+3} = \binom{18}{r-3} \] Using the property of combinations, the above equation can be rewritten as: \[\frac{18!}{(2r+3)!(18-(2r+3))!} = \frac{18!}{(r-3)!(18-(r-3))!} \] Cancel out \(18!\) from both sides: \[\frac{1}{(2r+3)!(15-2r)!} = \frac{1}{(r-3)!(21-r)!} \] Now, cross-multiply and simplify: \[((2r+3)!(15-2r)!) = ((r-3)!(21-r)!) \] At this point, we can perform manual checking for each possible value of r, given in the options (a), (b), (c), and (d). When r=6, LHS becomes: \[((2 \times 6 + 3)!(15 - 2 \times 6)!) = (15!)(3!) \] And RHS becomes: \[((6 - 3)!(21 - 6)!) = (3!)(15!) \] Since LHS = RHS if r = 6, the value of r that makes the coefficients of \((2 r+4)\)th term and the \((r-2)\)th term equal in the expansion of \((1+x)^{18}\) is: \(r = \boxed{6}\) The correct answer is (c).