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 being taken $11$ at a time and $c$ the number of permutations 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 total number of permutations of $n$things taken $r$at a time is given by, ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$. Use this formula to find $a,b$ and $c$.

Complete step-by-step answer:
We know that the number of permutations of $n$ things taken $r$ at a time is given by, ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$
Given, $a$ denotes the number of permutations of $x + 2$ things taken all at a time.
$\therefore a = {}^{x + 2}{P_{x + 2}} = \left( {x + 2} \right)!$ ….. (1)
$b$ denotes the number of permutations of $x$ things being taken $11$ at a time.
$\therefore b = {}^x{P_{11}} = \dfrac{{x!}}{{\left( {x - 11} \right)!}}$ …..(2)
$c$ denotes the number of permutations of $x - 11$ things taken all at a time.
$\therefore c = {}^{x - 11}{P_{x - 11}} = \left( {x - 11} \right)!$ ….. (3)
Given: $a = 182bc$
$\Rightarrow \left( {x + 2} \right)! = 182 \times \dfrac{{x!}}{{\left( {x - 11} \right)!}} \times \left( {x - 11} \right)!$ [Using (1),(2) and (3)]
$\Rightarrow \left( {x + 2} \right)! = 182x!$
$\Rightarrow \left( {x + 2} \right)\left( {x + 1} \right)x! = 182x!$ [Using $\left( {x + 2} \right)! = \left( {x + 2} \right)\left( {x + 1} \right)x!$]
$\Rightarrow \left( {x + 2} \right)\left( {x + 1} \right) = 182$
$\Rightarrow {x^2} + x + 2x + 2 = 182$
$\Rightarrow {x^2} + 3x - 180 = 0$ ….. (1)
Now using factorization method to find the roots of equation,
$\Rightarrow {x^2} + \left( {15 - 12} \right)x - 180 = 0$
$\Rightarrow {x^2} + 15x - 12x - 180 = 0$
$\Rightarrow x\left( {x + 15} \right) - 12\left( {x + 15} \right) = 0$
$\Rightarrow \left( {x + 15} \right)\left( {x - 12} \right) = 0$
$\Rightarrow x = - 15$ or $x = 12$
$\therefore x = 12$
Hence, option (B) is the correct answer.

Note: The roots of equation (1) can also be find by using the quadratic formula which is given by,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
On comparing ${x^2} + 3x - 180 = 0$ with $a{x^2} + bx + c = 0$, we get-
$a = 1,b = 3,c = - 180$
$\therefore x = \dfrac{{ - 3 \pm \sqrt {{{\left( 3 \right)}^2} - 4 \times 1 \times - 180} }}{{2 \times 1}}$
$\Rightarrow x = \dfrac{{ - 3 \pm \sqrt {9 + 720} }}{2}$
$\Rightarrow x = \dfrac{{ - 3 \pm \sqrt {729} }}{2}$
$\Rightarrow x = \dfrac{{ - 3 \pm 27}}{2}$
$\Rightarrow x = \dfrac{{ - 3 + 27}}{2}$ or $x = \dfrac{{ - 3 - 27}}{2}$
$\Rightarrow x = \dfrac{{24}}{2}$ or $x = \dfrac{{ - 30}}{2}$
$\Rightarrow x = 12$ or $x = - 15$
$\therefore x = 12$