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(B) Given that $a$ is the number of permutations of $x + 2$ things taken all at a time,so $a = (x + 2)!$.$b$ is the number of permutations of $x$ thin

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$12$

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(B) Given that $a$ is the number of permutations of $x + 2$ things taken all at a time,so $a = (x + 2)!$.$b$ is the number of permutations of $x$ things taken $11$ at a time,so $b = ^xP_{11} = \frac{x!}{(x - 11)!}$.$c$ is the number of permutations of $x - 11$ things taken all at a time,so $c = (x - 11)!$.Given the equation $a = 182bc$,we substitute the expressions:$(x + 2)! = 182 	imes \frac{x!}{(x - 11)!} 	imes (x - 11)!$$(x + 2)! = 182 	imes x!$$(x + 2)(x + 1)x! = 182x!$$(x + 2)(x + 1) = 182$$x^2 + 3x + 2 = 182$$x^2 + 3x - 180 = 0$$(x + 15)(x - 12) = 0$Since $x$ must be a positive integer such that $x \ge 11$ (from the definition of $^xP_{11}$),we have $x = 12$.

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

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