Binomial theorem question

$ \left( {3 + \frac{x} {2}} \right)^n = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right)3^k \left( {\frac{x} {2}} \right)^{n - k} } $

now we equate the aforementioned coefficients:

$ \left( {\begin{array}{*{20}c} n \\ {n - 7} \\ \end{array} } \right)3^{n - 7} \left( {\frac{1} {2}} \right)^7 = \left( {\begin{array}{*{20}c} n \\ {n - 8} \\ \end{array} } \right)3^{n - 8} \left( {\frac{1} {2}} \right)^8 $

$\begin{gathered} \Leftrightarrow 6\frac{{n!}} {{7!\left( {n - 7} \right)!}} = \frac{{n!}} {{8!\left( {n - 8} \right)!}} \hfill \\ \Leftrightarrow 48 = n - 7 \Rightarrow n = 55 \hfill \\ \end{gathered}$