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Question: If m parallel lines in a plane are intersected by a family of n parallel lines, the number of parall...

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. 14mn(m−1)(n−1)\dfrac{1}{4}mn\left( m-1 \right)\left( n-1 \right)
b. 12mn(m−1)(n−1)\dfrac{1}{2}mn\left( m-1 \right)\left( n-1 \right)
c. 14m2n2\dfrac{1}{4}{{m}^{2}}{{n}^{2}}
d. None of these

Explanation

Solution

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, nCr=n!r!(n−r)!^{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 nCr=n!r!(n−r)!^{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 mC2^{m}{{C}_{2}} and 2 out of n parallel lines as nC2^{n}{{C}_{2}}. Therefore, we can write the total number of parallelogram formed as, mC2×nC2^{m}{{C}_{2}}{{\times }^{n}}{{C}_{2}}. And by using the formula of combination, we can express it as,
m!2!(m−2)!×n!2!(n−2)! ⇒m(m−1)(m−2)!(2×1)(m−2)!×n(n−1)(n−2)!(2×1)(n−2)! \begin{aligned} & \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{aligned}
In the above expression, the common terms in the numerator and the denominator of both the terms will get cancelled out. So, we get,
m(m−1)2×n(n−1)2 ⇒14mn(m−1)(n−1) \begin{aligned} & \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{aligned}
Hence, the total number of parallelograms that can be formed is 14mn(m−1)(n−1)\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 nCr=n!r!(n−r)!^{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.