Chọn phương án C.

Điều kiện: \(\begin{cases}
x\in\mathbb{N}\\ x\geq1.
\end{cases}\)
$$\begin{eqnarray*}
&6\left(\mathrm{P}_x-\mathrm{P}_{x-1}\right)&=\mathrm{P}_{x+1}\\
\Leftrightarrow&6\left(x!-(x-1)!\right)&=(x+1)!\\
\Leftrightarrow&6\left(x(x-1)!-(x-1)!\right)&=(x+1)x(x-1)!\\
\Leftrightarrow&6(x-1)!\left(x-1\right)&=(x+1)x(x-1)!\\
\Leftrightarrow&6\left(x-1\right)&=(x+1)x\\
\Leftrightarrow&x^2-5x+6&=0\\
\Leftrightarrow&\left[\begin{array}{ll}x=2 &\text{(nhận)}\\ x=3 &\text{(nhận)}\end{array}\right.
\end{eqnarray*}$$