If nP3 - 6 nC4 0 find the value of n - JAMB Mathematics 2003 Question

\(^{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)!\((n-4)!4!)=6(n-3)!\\frac{(n-4)!}{(n-3)!}=\frac{6}{4!}\\frac{(n-4)!}{(n-3)(n-4)!}=\frac{6}{4 \times 3\times 2\times 1}\\frac{1}{(n-3)}=]\frac{1}{4}\n-3=4\n=4+3\n=7\)