Question

If p denotes the number of permutations of $(x+3)$ things taken all at a time, q denotes the number of permutations of $(x+1)$ things taken 11 at a time and r denotes denotes the number permutations of $(x-10)$ things taken all at a time such that $p=182qr,$ then the value of x is

• A. $15$
• B. $11$
• C. $12$
• D. $10$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $15$
B $11$
C $12$
D $10$

Given that $p$ denotes the number of permutations of $(x+3)$ things all at a time.

$\Rightarrow p{=}^{x+3}{P}_{x+3}=(x+3)!$

$q$ denotes the number of permutations of $(x+1)$ things taken 11 at a time.

$\Rightarrow q{=}^{x+1}{P}_{11}=\frac{(x+1)!}{\left(\right(x+1)-11)!}$

$r$ denotes the number of permutations of $(x-10)$ things taken 11 at a time.

$\Rightarrow r{=}^{x-10}{P}_{x-10}=(x-10)!$

Given that,

$p=182qr$

Put the values of p, q and r, we get,

$(x+3)!=182\times \frac{(x+1)!}{(x-10)!}\times (x-10)!$

$\Rightarrow (x+3)!=182\times (x+1)!$

$(x+3)(x+2)(x+1)!=182\times (x+1)!$

$(x+3)(x+2)=182$

$x^2+5x+6=182$

$x^2+5x-176=0$

$(x-11)(x+16)=0$

$x=11,-16$

Values of x $=11,-16$