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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 $a=182 b c$, then the value of $x$ is
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The correct answer is:
$12$
We have $a={ }^{x+2} P_{x+2}=(x+2)!, b={ }^x P_{11}=\frac{x!}{(x-11)!}$
and $c={ }^{x-11} P_{x-11}=(x-11)!$
Now $a=182 b c \Rightarrow(x+2)!=182 \cdot \frac{x!}{(x-11)!}(x-11)!$
$\Rightarrow(x+2)!=182 x!\Rightarrow(x+2)(x+1)=182 \Rightarrow x=12$.
and $c={ }^{x-11} P_{x-11}=(x-11)!$
Now $a=182 b c \Rightarrow(x+2)!=182 \cdot \frac{x!}{(x-11)!}(x-11)!$
$\Rightarrow(x+2)!=182 x!\Rightarrow(x+2)(x+1)=182 \Rightarrow x=12$.
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