If \(^{n}P_{3} - 6(^{n}C_{4}) = 0\), find the value of n.
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Correct Answer: Option C
Explanation:
\(^{n}P_3 - 6(^{n}C_{4})=0\\\frac{n!}{(n-3)!}-6\left( \frac{n!}{(n-4)!4!}\right)=0\\\frac{n!}{(n-3)!}=6\left(\frac{n!}{(n-4)!4!}\right)\\n!((n-4)!4!)=6n!(n-3)!\\((n-4)!4!)=6(n-3)!\\\frac{(n-4)!}{(n-3)!}=\frac{6}{4!}\\\frac{(n-4)!}{(n-3)(n-4)!}=\frac{6}{4 \times 3\times 2\times 1}\\\frac{1}{(n-3)}=]\frac{1}{4}\\n-3=4\\n=4+3\\n=7\)
\(^{n}P_3 - 6(^{n}C_{4})=0\\\frac{n!}{(n-3)!}-6\left( \frac{n!}{(n-4)!4!}\right)=0\\\frac{n!}{(n-3)!}=6\left(\frac{n!}{(n-4)!4!}\right)\\n!((n-4)!4!)=6n!(n-3)!\\((n-4)!4!)=6(n-3)!\\\frac{(n-4)!}{(n-3)!}=\frac{6}{4!}\\\frac{(n-4)!}{(n-3)(n-4)!}=\frac{6}{4 \times 3\times 2\times 1}\\\frac{1}{(n-3)}=]\frac{1}{4}\\n-3=4\\n=4+3\\n=7\)