Question

If $a$ represents the number of permutations of $(x+2)$ things taken together, $b$ represents the number of permutations of $11$ things taken together out of $x$ things, and $c$ represents the number of permutations of $(x-11)$ things taken together so that $a=182bc$, then $x=$

• A. $15$
• B. $12$
• C. $10$
• D. $18$

Answer 1

Answer: (B)

$12$

$a{=}^{x+2}{P}_{x+2},b{=}^x{P}_{11},c{=}^{x-11}{P}_{x-11}$

$\Rightarrow a=(x+2)!,b=\frac{x!}{(x-11)!},c=(x-11)!$

Now $a=182bc$

$\Rightarrow (x+2)!=182\frac{x!}{(x-11)!}(x-11)!\Rightarrow (x+2)(x+1)x!=182x!\Rightarrow (x+2)(x+1)=182=14\times 13\Rightarrow x+1=13\Rightarrow x=12$