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Let $a_{1}, a_{2}, \ldots a_{n}$ be $n$ nonzero real numbers, of which $p$ are positive and remaining are negative. The number of ordered pairs $(\mathrm{j}, \mathrm{k}), \mathrm{j} < \mathrm{k}$, for which $\mathrm{a}_{\mathrm{j}} \mathrm{a}_{\mathrm{k}}$ is positive, is $55 .$ Similarly, the number of ordered pairs $(\mathrm{j}, \mathrm{k}), \mathrm{j} < \mathrm{k}$, for which a $a_{k}$ is negative is 50 . Then the value of $p^{2}+(n-p)^{2}$ is
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125
${ }^{\mathrm{P} C}_{2}+{ }^{n-p} \mathrm{C}_{2}=55$
$\frac{\mathrm{p}(\mathrm{p}-1)}{2}+\frac{(\mathrm{n}-\mathrm{p})(\mathrm{n}-\mathrm{p}-1)}{2}=55$ .........(i)
Also,$p(n-p)=50$.........(ii)
Put in (i)
$\begin{array}{l}
p(p-1)+\frac{50}{p}\left(\frac{50}{p}-1\right)=110 \\
p^{2}-p+\left(\frac{50}{p}\right)^{2}-\frac{50}{p}=110 \\
p^{2}-p+\left(\frac{50}{p}\right)^{2}-\frac{50}{p}=110 \\
\left(p+\frac{50}{p}\right)^{2}-100-\left(p+\frac{50}{p}\right)=110 \\
t^{2}-t-210=0 \\
t=15 \text { or }-14 \text { (not true) } \\
\therefore \quad p+\frac{50}{p}=15
\end{array}$
$\therefore$ To find $p^{2}+(n-p)^{2}=p^{2}+\left(\frac{50}{p}\right)^{2}$
$\begin{array}{l}
=\left(p+\frac{50}{p}\right)^{2}-100 \\
=125 \text { (using (iii)) }
\end{array}$
$\frac{\mathrm{p}(\mathrm{p}-1)}{2}+\frac{(\mathrm{n}-\mathrm{p})(\mathrm{n}-\mathrm{p}-1)}{2}=55$ .........(i)
Also,$p(n-p)=50$.........(ii)
Put in (i)
$\begin{array}{l}
p(p-1)+\frac{50}{p}\left(\frac{50}{p}-1\right)=110 \\
p^{2}-p+\left(\frac{50}{p}\right)^{2}-\frac{50}{p}=110 \\
p^{2}-p+\left(\frac{50}{p}\right)^{2}-\frac{50}{p}=110 \\
\left(p+\frac{50}{p}\right)^{2}-100-\left(p+\frac{50}{p}\right)=110 \\
t^{2}-t-210=0 \\
t=15 \text { or }-14 \text { (not true) } \\
\therefore \quad p+\frac{50}{p}=15
\end{array}$
$\therefore$ To find $p^{2}+(n-p)^{2}=p^{2}+\left(\frac{50}{p}\right)^{2}$
$\begin{array}{l}
=\left(p+\frac{50}{p}\right)^{2}-100 \\
=125 \text { (using (iii)) }
\end{array}$
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