Question

If $a$ denotes the no. of permutation of $x+2$ things taken all at a time, $b$ the no. of permutation of $x$ things taken $11$ at a time and $c$ the no. of permutation of $x–11$ things taken all at a time such that $a=182bc$ then the value of $x$ is?

• A. $15$
• B. $12$
• C. $10$
• D. $18$

Answer 1

The correct option is B

$12$

Explanation for the correct option:

Step 1: Find the value of $a,b\mathrm{and}c$

Given,

$\begin{array}{rcl}a& =& {}^{x+2}P_{x+2}\\ & =& \left(x+2\right)!\end{array}$ $\left[\mathbf{\because }{}^{\mathbf{n}}\mathit{P}_{\mathbf{n}}\mathbf{}\mathbf{=}\mathbf{}\mathit{n}\mathbf{!}\right]$

$\begin{array}{rcl}b& =& {}^{x}P_{11}\\ & =& \frac{x!}{(x-11)!}\end{array}$

$\begin{array}{rcl}c& =& {}^{x-11}P_{x-11}\\ & =& \left(x-11\right)!\end{array}$

Step 2: Find the value of $x$

It is given that, $a=182bc$

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

$\Rightarrow$ $(x+2)!=182x!$

$\Rightarrow$ $(x+2)(x+1)x!=182x!$

$\Rightarrow$ $(x+2)(x+1)=182$

$\Rightarrow$ $x^2+3x+2=182$

$\Rightarrow$ $x^2+3x-180=0$

$\Rightarrow$$x^2+15x-12x-180=0$

$\Rightarrow$$x(x+15)-12(x+15)=0$

$\Rightarrow$$(x+15)(x-12)=0$

$\mathbf{\therefore }\mathbf{}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{12}\mathbf{,}\mathbf{-}\mathbf{15}$

Hence, Option ‘B’ is Correct.