Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal 

Question: Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal.



My attempt:
$\displaystyle \binom{n}{7}=\binom{n}{8} $
$$ n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) \times 2^{n-7} \times (\frac{1}{3})^7= n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7) \times 2^{n-8} \times (\frac{1}{3})^8 $$
$$ \frac{6}{7!} = \frac{n-7}{40320} $$
$$ n-7 = 48 $$
$$ n=55 $$

The coefficient of $x^7$ will be ${n\choose 7}{2^{n-7}\over 3^7}$ and the coefficient of $x^8$ will be ${n\choose 8}{2^{n-8}\over 3^8}$ and not what you have got .you just missed out the factor of 3 that plays a role as a coefficient.So equating it you get
$${n\choose 7}{2^{n-7}\over 3^7}={n\choose 8}{2^{n-8}\over 3^8}$$
and get the answer as $55$
