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 $a = 182\ bc$, then the value of $x$ is

Question

Dr. Abubacker · Accepted Answer

(b)
We have, $a = \ ^{x + 2}P_{x + 2} = (x + 2)!$ ,
and $b = \ ^{x}P_{11} = \frac{x!}{(x - 11)!}$
and $c = \ ^{x - 11}P_{x - 11} = (x - 11)!$
Now, $a = 182\ bc$
$	herefore\ (x + 2)! = 182.\frac{x!}{(x - 11)!}(x - 11)!$
$\Rightarrow (x + 2)! = 182\ x!$
$\Rightarrow (x + 2)(x + 1) = 182$
$\Rightarrow x^{2} + 3x - 180 = 0$
$\Rightarrow (x - 12)(x + 15) = 0$
$\Rightarrow x = 12,\  - 15$
$	herefore$ Neglect the negative value of $x$.
$\Rightarrow \ x = 12$

Q. 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 $a = 182\ bc$, then the value of $x$ is

Explanation

Reviewed by Dr. Abubacker

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