Question

If a denotes the number of permutations of (x + 2) things taken all at a time, b the number of permutations of x things taken 11 at a time and c the number of permutations of x − 11 things taken all at a time such that a = 182 bc, find the value of x.

Answer 1

a = ^x+2P_x+2 = (x+2)!
b = ^xP₁₁ = $\frac{x!}{(x-11)!}$
c = ^x$-$11P_x$-$11 = $(x-11)!$

a = 182 bc
$\Rightarrow$ (x+2)! = 182 $\frac{x!}{\left(x-11\right)!}$ $\times \left(x-11\right)!$
$\Rightarrow$(x+2)! = 182 (x!)

$$\begin{gathered} \Rightarrow \frac{\left(x+2\right)!}{x!}=182 \\ \Rightarrow \left(x+2\right)\left(x+1\right)=182 \\ \Rightarrow \left(x+2\right)\left(x+1\right)=14\times 13 \\ \Rightarrow x+2=14 \\ \Rightarrow x=12 \end{gathered}$$