Explanation

solution $$\begin{gathered} {}^n{P}_3-6{(}^n{C}_4)=0 \\ \frac{n!}{(n-3)!}-6\left(\frac{n!}{(n-4)!4!}\right)=0 \\ \frac{n!}{(n-3)!}=6\left(\frac{n!}{(n-4)!4!}\right) \\ n!((n-4)!4!)=6n!(n-3)! \\ ((n-4)!4!)=6(n-3)! \\ \frac{(n-4)!}{(n-3)!}=\frac{6}{4!} \\ \frac{(n-4)!}{(n-3)(n-4)!}=\frac{6}{4\times 3\times 2\times 1} \\ \frac{1}{(n-3)}=]\frac{1}{4} \\ n-3=4 \\ n=4+3 \\ n=7 \end{gathered}$$

solution $$\begin{gathered} {}^n{P}_3-6{(}^n{C}_4)=0 \\ \frac{n!}{(n-3)!}-6\left(\frac{n!}{(n-4)!4!}\right)=0 \\ \frac{n!}{(n-3)!}=6\left(\frac{n!}{(n-4)!4!}\right) \\ n!((n-4)!4!)=6n!(n-3)! \\ ((n-4)!4!)=6(n-3)! \\ \frac{(n-4)!}{(n-3)!}=\frac{6}{4!} \\ \frac{(n-4)!}{(n-3)(n-4)!}=\frac{6}{4\times 3\times 2\times 1} \\ \frac{1}{(n-3)}=]\frac{1}{4} \\ n-3=4 \\ n=4+3 \\ n=7 \end{gathered}$$