Question

The value of $\lim_{n \to \infty} \frac{{}^nC_3 - {}^nP_3}{n^3}$ is equal to:

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

Hint:

Remember that ${}^nC_r = \frac{{}^nP_r}{r!}$. For the highest power $n^r$, the coefficient in ${}^nC_r$ is $\frac{1}{r!}$ and in ${}^nP_r$ it is $1$. Here, $(\frac{1}{6} - 1) = -\frac{5}{6}$.

Answer 1

The Correct Option is A

Concept: Expand the combination and permutation formulas to express them as polynomials in $n$. For limits as $n \to \infty$ of rational functions, the result is the ratio of the coefficients of the highest power of $n$.
• ${}^nC_3 = \frac{n(n-1)(n-2)}{3!}$
• ${}^nP_3 = n(n-1)(n-2)$

Step 1: Expand the numerator.
\[ {}^nC_3 - {}nP_3 = \frac{n(n-1)(n-2)}{6} - n(n-1)(n-2) \] Factor out $n(n-1)(n-2)$: \[ n(n-1)(n-2) \left( \frac{1}{6} - 1 \right) = n(n-1)(n-2) \left( -\frac{5}{6} \right) \]

Step 2: Find the highest power of $n$.
The expression $n(n-1)(n-2)$ is a third-degree polynomial: \[ n(n^2 - 3n + 2) = n^3 - 3n^2 + 2n \] So the numerator is $-\frac{5}{6}n^3 + \frac{15}{6}n^2 - \dots$

Step 3: Apply the limit $n \to \infty$.
\[ \lim_{n \to \infty} \frac{-\frac{5}{6}n^3 + \frac{15}{6}n^2 - \dots}{n^3} = -\frac{5}{6} \]