Question

If a denotes the number of permutations of (x+2) things taken all at a time, b the number of permutations of x-11 things taken all at a time such that a = 182 bc, find the value of x.

Answer 1

'a' denotes the number of permutations of (x+2) things taken all at a time.
'b' is the number of permutations of x things taken 11 at a time.
$\therefore b=x_{P_{11}}$ and, C is the number of permutations of x-11 things taken all at a time.
$\therefore C={x-11}_{P_{x-11}}$
Now,
a = 182 bc [given]
$\Rightarrow {x+2}_{P_{x+2}}=182\times x_{P_{11}}$
$\Rightarrow (x+2)!=182\times \frac{x!}{(x-11)!}\times (x-11)!$
$\left[\begin{array}{c}\because n_{P_n}=n!\\ and{n}_{P_r}=\frac{n!}{(n-r)!}\end{array}\right]$
$\Rightarrow (x+2)!=182\times x!\Rightarrow (x+2)(x+1)x!=182\times x!\Rightarrow (x+2)(x+1)=182\Rightarrow x^2+3x+2-182=0\Rightarrow x^2+3x-180=0\Rightarrow x^2+15x-12x-180=0\Rightarrow x(x+15)-12(x+15)=0\Rightarrow (x-12)(x+15)=0\Rightarrow x-12=0[\because x\ne -15]\Rightarrow x=12$
Hence, x= 12