Question

The value of $\underset{n\to \infty }{\lim }\frac{{}^n{c}_3{-}^n{P}_3}{n^3}$ is equal to

• A. $\frac{-5}{6}$
• B. $\frac{5}{6}$
• C. $\frac{1}{6}$
• D. $-\frac{1}{6}$
• E. $\frac{2}{3}$

Answer 1

Answer: (A)

We have, ${\lim }_{n\to \infty }\left[\frac{{}^n{C}_3-{}^n{P}_3}{n^3}\right]$
$=\underset{n\to \infty }{\lim }\left[\frac{{}^n{C}_3-{}^n{C}_{3}3!}{n^3}\right]$
$=\underset{n\to \infty }{\lim }\left[\frac{{}^n{C}_3[1-3!]}{n^3}\right]$
$=-5\underset{n\to \infty }{\lim }\left[\frac{n(n-1)(n-2)}{1\cdot 2\cdot 3\cdot n^3}\right]$
$=-\frac{5}{6}\underset{n\to \infty }{\lim }\left[1\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\right]=-\frac{5}{6}$