Question

If $P_n=C_0.C_1.C_2.....C_n$ and $8!\cdot P_{n+1}=9^8\cdot P_n,$ then the value of $n$ is :

• A. $8$
• B. $9$
• C. $10$
• D. None of these

Answer 1

Answer: (A)

$8$

$\frac{P_{n+1}}{P_n}=\frac{{}^{(n+1)}{C}_0{.}^{(n+1)}{C}_1....{.}^{(n+1)}{C}_{n+1}}{{}^n{C}_0{.}^n{C}_1...{.}^n{C}_n}$
$=\frac{\left(1\right)}{\left(1\right)}\frac{{}^{(n+1)}{C}_1}{{}^n{C}_1}.....\frac{{}^{(n+1)}{C}_n}{{}^n{C}_n}.1$
$=\frac{n+1}{n}.\frac{(n+1)}{n-1}.\frac{n+1}{n-2}$ .... $\frac{n+1}{1}$
$=\frac{(n+1{)}^n}{\lfloor n}$
But given $\frac{P_{n+1}}{P_n}=\frac{9^8}{8!}\Rightarrow n=8$