Correct Answer: Option D

Explanation

\(\frac{^{n}P_{3}}{^{n}C_{2}} + ^{n}P_{0}\)

\(\frac{^{n}P_{3}}{^{n}C_{2}} = \frac{n!}{(n - 3)!} ÷ \frac{n!}{(n - 2)! 2!}\)

\(\frac{n!}{(n - 3)!} \times \frac{(n - 2)(n - 3)! 2!}{n!} = 2n - 4\)

\(^{n}P_{0} = \frac{n!}{(n - 0)!} = 1\)

\(\frac{^{n}P_{3}}{^{n}C_{2}} + ^{n}P_{0} = 2n - 4 + 1 = 2n - 3\)

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