Question

Let $a_1,a_2,....a_n$ be $n$ nonzero real numbers, of which $p$ are positive and remaining are negative. The number of ordered pairs $(j,k),j<k$, for which $a_j{a}_k$ is positive, is 55. Similarly, the number of ordered pairs $(j,k),j<k$, for which $a_j{a}_k$ is negative is 50. Then the value of $p^2+(n–p{)}^2$ is

• A. 629
• B. 325
• C. 125
• D. 221

Answer 1

The correct option is C 125

There are $p$ positive numbers and $(n-p)$ negative numbers.
The number of pairs for $a_j{a}_k$ to be negative,
${}^p{C}_1{\times }^{(n-p)}{C}_1=50$
$\Rightarrow p(n-p)=50$..........(i)
The number of pairs for $a_j{a}_k$ to be positive,
${}^p{C}_2{+}^{(n-p)}{C}_2=55$
$p(p-1)+(n-p)(n-p-1)=110$
using equation (i),
$p(p-1)+\frac{50}{p}(\frac{50}{p}-1)=110\Rightarrow p^2-p+{\left(\frac{50}{p}\right)}^2-\left(\frac{50}{p}\right)=110\Rightarrow {(p+\frac{50}{p})}^2-(p+\frac{50}{p})-100=110$
Assuming the $(p+\frac{50}{p})=y$
$\Rightarrow y^2-y-210=0\Rightarrow (y-15)(y+14)=0$
$\therefore y=15\Rightarrow (p+\frac{50}{p})=15$
[$\because y$ cannot be negative]

$\therefore p^2+(n-p{)}^2=p^2+{\left(\frac{50}{p}\right)}^2={(p+\frac{50}{p})}^2-100=125$