Find $P^n_2$ if the fifth term of the expansion of $\left(\sqrt[3]{x} + \dfrac{1}{x}\right)^n$ does not depend on $x$.
Fifth term of the expansion of $\left(\sqrt[3]{x} + \dfrac{1}{x}\right)^n$ is
$T_5 = T_{4+1} = C^n_4 \; \left(x^{\frac{1}{3}}\right)^{n-4} \; \left(x^{-1}\right)^4$
i.e. $\;$ $T_5 = C^n_4 \; x^{\left(\frac{n}{3} - \frac{4}{3} - 4\right)}$
i.e. $\;$ $T_5 = C^n_4 \; x^{\left(\frac{n-16}{3}\right)}$
As per question, the fifth term in the binomial expansion is independent of $x$.
$\implies$ $\dfrac{n-16}{3} = 0$ $\implies$ $n = 16$
Now, $\;$ $P^n_2 = P^{16}_2 = \dfrac{16!}{\left(16 - 2\right)!} = \dfrac{16 \times 15 \times 14!}{14!} = 240$