How many positive terms are there in the sequence $\left(x_n\right)$ if $x_n = \dfrac{195}{4 \times P_n} - \dfrac{P^{n+3}_{3}}{P_{n+1}}$, $\;\;\;$ $n \in N$
$\begin{aligned}
x_n & = \dfrac{195}{4 \times P_n} - \dfrac{P^{n+3}_{3}}{P_{n+1}} \\\\
& = \dfrac{195}{4 \times n!} - \dfrac{\left(n+3\right)!}{\left(n+3-3\right)! \left(n+1\right)!} \\\\
& = \dfrac{195}{4 \times n!} - \dfrac{\left(n+3\right) \left(n+2\right) \left(n+1\right)!}{n! \left(n+1\right)!} \\\\
& = \dfrac{195}{4 \times n!} - \dfrac{\left(n+3\right) \left(n+2\right)}{n!} \\\\
& = \dfrac{195 - 4 \left(n^2 + 5n + 6\right)}{4 \times n!} \\\\
& = \dfrac{195 - 4n^2 - 20n - 24}{4 \times n!} \\\\
& = \dfrac{171 - 20n - 4n^2}{ 4 \times n!}
\end{aligned}$
When $\;$ $n = 1$, $\;$ $x_n = \dfrac{171 - 20 - 4}{4 \times 1!} = \dfrac{147}{4 \times 1} > 0$
When $\;$ $n = 2$, $\;$ $x_n = \dfrac{171 - 40 - 16}{4 \times 2!} = \dfrac{115}{4 \times 2} > 0$
When $\;$ $n = 3$, $\;$ $x_n = \dfrac{171 - 60 - 36}{4 \times 3!} = \dfrac{75}{4 \times 6} > 0$
When $\;$ $n = 4$, $\;$ $x_n = \dfrac{171 - 80 - 64}{4 \times 4!} = \dfrac{27}{4 \times 24} > 0$
When $\;$ $n = 5$, $\;$ $x_n = \dfrac{171 - 100 - 100}{4 \times 5!} = \dfrac{-29}{4 \times 120} < 0$
$\therefore \;$ There are $4$ positive terms in the given sequence.