n - 5
n - 3
2n - 1
2n - 3
Correct answer is D
\(\frac{^{n}P_{3}}{^{n}C_{2}} + ^{n}P_{0}\)
\(\frac{^{n}P_{3}}{^{n}C_{2}} = \frac{n!}{(n - 3)!} ÷ \frac{n!}{(n - 2)! 2!}\)
\(\frac{n!}{(n - 3)!} \times \frac{(n - 2)(n - 3)! 2!}{n!} = 2n - 4\)
\(^{n}P_{0} = \frac{n!}{(n - 0)!} = 1\)
\(\frac{^{n}P_{3}}{^{n}C_{2}} + ^{n}P_{0} = 2n - 4 + 1 = 2n - 3\)