If the coefficient of${\left(2r+4\right)}^{th}$  term and${\left(r-2\right)}^{th}$  terms in the expansion of

${\left(1+x\right)}^{18}$ are equal then $r=$

Given expansion ${\left(1+x\right)}^{18}$ is

 We have general term in the expansion ${\left(x+a\right)}^n$

$(\therefore T_{r+1}={}^n{\text{C}}_r{x}^{n-r}{\left(a\right)}^r$ be the expansion of ${(x+a)}^n)$

${\left(2r+4\right)}^{th}$  term:

$T_{r+1}={}^n{\text{C}}_r{x}^{n-r}{\left(a\right)}^r$

$T_{\left(2r+3\right)+1}={}^{18}{\text{C}}_{2r+3}{\left(1\right)}^{18-2r-3}{\left(x\right)}^{2r+3}$  

The coefficient of${\left(2r+4\right)}^{th}$  term is ${}^{18}{\text{C}}_{2r+3}$

${\left(r-2\right)}^{th}$  term:

$T_{\left(r+1\right)+1}={}^n{\text{C}}_{r+1}{x}^{n-r-1}{\left(a\right)}^{r+1}$

$T_{\left(r-3\right)+1}={}^{18}{\text{C}}_{r-3}{\left(1\right)}^{18-r+3}{\left(x\right)}^{r-3}$  

The coefficient of${\left(r-2\right)}^{th}$  term is ${}^{18}{\text{C}}_{r-3}$

 The coefficient of${\left(2r+4\right)}^{th}$  term and${\left(r-2\right)}^{th}$  terms in the expansion of

${\left(1+x\right)}^{18}$ are equal then

$\Rightarrow T_{2r+4}=T_{r-2}$

                ${}^{18}{\text{C}}_{2r+3}{=}^{18}{\text{C}}_{r-3}$  $\left({\therefore }^n{\text{C}}_{2r}{=}^n{\text{C}}_{r-1}\Rightarrow 2r=r-1,n=2r+r-1\right)$

                $18=2r+3+r-3$

                      $\because r=6$