Question

The value of $\displaystyle\lim_{n\to\infty} \frac{^{n}c_{3}-^{n}P_{3}}{n^{3}}$ 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, $\lim _{n \rightarrow \infty}\left[\frac{{ }^{n} C_{3}-{ }^{n} P_{3}}{n^{3}}\right]$
$=\displaystyle\lim _{n \rightarrow \infty}\left[\frac{{ }^{n} C_{3}-{ }^{n} C_{3} 3 !}{n^{3}}\right]$
$=\displaystyle\lim _{n \rightarrow \infty}\left[\frac{{ }^{n} C_{3}[1-3 !]}{n^{3}}\right]$
$=-5 \displaystyle\lim _{n \rightarrow \infty}\left[\frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3 \cdot n^{3}}\right]$
$=-\frac{5}{6} \displaystyle\lim _{n \rightarrow \infty}\left[1\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\right]=-\frac{5}{6}$