Question

If m parallel lines in a plane are intersected by a family of n parallel lines, the number of parallelogram that can be formed is
a. $\dfrac{1}{4}mn\left( m-1 \right)\left( n-1 \right)$
b. $\dfrac{1}{2}mn\left( m-1 \right)\left( n-1 \right)$
c. $\dfrac{1}{4}{{m}^{2}}{{n}^{2}}$
d. None of these

Answer 1

Hint: To solve this question, we should know that a parallelogram is formed of 2 pairs of parallel lines and to choose the 2 pairs of parallel lines, we will use the formula of combination which is used to choose r out of n items irrespective of their orders, that is, $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. By using these concepts we will solve this question.

Complete step-by-step answer:
In this question, we have been asked to find the number of parallelograms that can be formed when m parallel lines in a plane are intersected by a family of n parallel lines. To solve this question, we should know that when we have to choose r out of n items irrespective of their orders, so we will apply the formula of combination, that is $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Now, we know that a parallelogram is formed by 2 pairs of parallel lines. So, to form the parallelogram from a given set of lines, we will choose 2 lines from m parallel lines and the other 2 from n parallel lines. So, by using the formula of combination, we can choose 2 out of m parallel lines as $^{m}{{C}_{2}}$ and 2 out of n parallel lines as $^{n}{{C}_{2}}$. Therefore, we can write the total number of parallelogram formed as, $^{m}{{C}_{2}}{{\times }^{n}}{{C}_{2}}$. And by using the formula of combination, we can express it as,
$\begin{align}
  & \dfrac{m!}{2!\left( m-2 \right)!}\times \dfrac{n!}{2!\left( n-2 \right)!} \\
 & \Rightarrow \dfrac{m\left( m-1 \right)\left( m-2 \right)!}{\left( 2\times 1 \right)\left( m-2 \right)!}\times \dfrac{n\left( n-1 \right)\left( n-2 \right)!}{\left( 2\times 1 \right)\left( n-2 \right)!} \\
\end{align}$
In the above expression, the common terms in the numerator and the denominator of both the terms will get cancelled out. So, we get,
$\begin{align}
  & \dfrac{m\left( m-1 \right)}{2}\times \dfrac{n\left( n-1 \right)}{2} \\
 & \Rightarrow \dfrac{1}{4}mn\left( m-1 \right)\left( n-1 \right) \\
\end{align}$
Hence, the total number of parallelograms that can be formed is $\dfrac{1}{4}mn\left( m-1 \right)\left( n-1 \right)$.
Therefore, option (a) is the correct answer.

Note: While solving this question, the possible mistake one can make is by writing the wrong formula for combination. The correct formula is $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Also, we can solve this question by assuming the values of m and n but that can be a confusing method. So it is better to refer to the combination method.