Question

Let be $a_1,a_2,....,a_n$ nonzero $n$ real numbers, $p$ of which 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 $125$

Given $a_1,a_2,.....a_n$ are non zero $n$ real numbers and also given $p$ of them are positive.

Now from above given data

$⟹$ Number of ordered pairs $(j,k),j<k$ for which $a_j.a_k$ is positive is 55. So either both of them are positive or both of them negative is the only way such that product is positive.

$⟹$ $\frac{p(p-1)}{2}+\frac{(n-p)(n-p-1)}{2}=55$ Combinations of two numbers from each set of positive and negative numbers

From above we get $p^2+{(n-p)}^2-n=110$ ----------- $\left(1\right)$

$⟹$ Number of ordered pairs $(j,k),j<k$ for which $a_j.a_k$ is negative is 50. So only possible case is one number is positive and other is negative.

It can be calculated as difference between all the combinations of products and products which are positive, so that remaining pairs have negative product

$⟹$ $\frac{n(n-1)}{2}-55=50$ as number of positive product pairs are $55$ which gives values of $n$ as $15,-14$

Therefore as $n$ is positive, we get value of $n=15$.

From equation $\left(1\right)$, we get $p^2+{(n-p)}^2=125$