If \(\frac{^{n}C_{3}}{^{n}P_{2}} = 1\), find the value of n.
A. 8
B. 7
C. 6
D. 5
Correct Answer: A
Explanation
\(^{n}C_{3} = \frac{n!}{(n - 3)! 3!}\)
\(^{n}P_{2} = \frac{n!}{(n - 2)!}\)
\(\frac{^{n}C_{3}}{^{n}P_{2}} = \frac{n!}{(n - 3)! 3!} ÷ \frac{n!}{(n - 2)!}\)
\(\frac{n!}{(n - 3)! 3!} \times \frac{(n - 2)!}{n!} = \frac{(n - 2)!}{(n - 3)! 3!}\)
Note that \((n - 2)! = (n - 2) \times (n - 2 - 1)! = (n - 2)(n - 3)!\)
\(\frac{(n - 2)(n - 3)!}{(n - 3)! 3!} = 1\)
\(\frac{n - 2}{3!} = 1 \implies n - 2 = 6\)
\(n = 2 + 6 = 8\)
