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 $a = 182\mspace{6mu} bc$, then the value of $x$ is.

• A. 15
• B. 12
• C. 10
• D. 18

Answer 1

Answer: (B)

12

Answer 2

We have $a =^{x + 2} ⥂ P_{x + 2} = (x + 2) ⥂ \mspace{6mu}!,\mspace{6mu} b =^{x} ⥂ P_{11} = \frac{x\mspace{6mu}!}{(x - 11)\mspace{6mu}!}$

and $c =^{x - 11} ⥂ P_{x - 11} = (x - 11)\mspace{6mu}!$

Now $a = 182\mspace{6mu} bc \Rightarrow (x + 2)\mspace{6mu}! = 182\mspace{6mu}.\mspace{6mu}\frac{x ⥂ \mspace{6mu}!}{(x - 11)\mspace{6mu}!}(x - 11)\mspace{6mu}!$

$\Rightarrow \mspace{6mu}(x + 2)!\mspace{6mu} = 182x\mspace{6mu}! \Rightarrow (x + 2)(x + 1) = 182 \Rightarrow x = 12$.