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

The correct option is A $\frac{-5}{6}$

We have
$\underset{n\to \infty }{lim}\left[\frac{{{}^{n}C}_3-{{}^{n}P}_3}{n^3}\right]$

$=\underset{n\to \infty }{lim}\left[\frac{{{}^{n}C}_3-{{}^{n}C}_{3}3!}{n^3}\right]=\underset{n\to \infty }{lim}\left[\frac{{{}^{n}C}_3-[1-3!]}{n^3}\right]$

$=-5\underset{n\to \infty }{lim}\left[\frac{n(n-1)(n-2)}{{\mathrm{1.2.3.}n}^3}\right]=-\frac{5}{6}\underset{n\to \infty }{lim}\left[1(1-\frac{1}{n})(1-\frac{2}{n})\right]=-\frac{5}{6}$