Question

The number of parallelograms that can be formed from a set of four parallel lines intersecting three parallel lines is:
(a) 6
(b) 18
(c) 12
(d) 9

Answer 1

Hint: We know that a parallelogram is constructed by using two parallel lines intersecting another two parallel lines. As in the question we are asked to find how many parallelograms could be possible from a set of four parallel lines intersecting three parallel lines so we can select two parallel lines from four parallel lines i.e. ${}^{4}{{C}_{2}}$ and multiplying this by selecting two parallel lines from another three parallel lines intersecting the four lines i.e. ${}^{3}{{C}_{2}}$ will get us the required answer.

Complete step-by-step answer:
We know that a parallelogram is constructed using two parallel lines intersecting other pairs of parallel lines so basically we require 4 lines to construct a parallelogram. In the below figure, you can see a parallelogram ABCD.

In the question above, we have to find the possible number of parallelograms from 4 parallel lines intersecting three parallel lines so implementing how a parallelogram is constructed we first of select 2 parallel lines from the given 4 parallel lines (using the combination formula ${}^{4}{{C}_{2}}$) and then multiply this result with the selection of two parallel lines from 3 intersecting parallel lines (using the combination formula ${}^{3}{{C}_{2}}$).
Number of possible parallelograms from a set of 4 parallel lines intersecting 3 parallel lines is:
${}^{4}{{C}_{2}}\times {}^{3}{{C}_{2}}$
We know that:
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Using the above formula in finding the expansion of ${}^{4}{{C}_{2}}\And {}^{3}{{C}_{2}}$ we get,
$\begin{aligned} & {}^{4}{{C}_{2}}=\dfrac{4!}{2!\left( 4-2 \right)!} \\\ & \Rightarrow {}^{4}{{C}_{2}}=\dfrac{4!}{2!\left( 2 \right)!} \\\ & \Rightarrow {}^{4}{{C}_{2}}=\dfrac{4.3.2!}{2!\left( 2 \right)!} \\\ & \Rightarrow {}^{4}{{C}_{2}}=\dfrac{4.3}{2.1}=6 \\\ \end{aligned}$
Expanding ${}^{3}{{C}_{2}}$ we get,
$\begin{aligned} & {}^{3}{{C}_{2}}=\dfrac{3!}{2!\left( 3-2 \right)!} \\\ & \Rightarrow {}^{3}{{C}_{2}}=\dfrac{3!}{2!\left( 1 \right)!} \\\ & \Rightarrow {}^{3}{{C}_{2}}=\dfrac{3.2!}{2!} \\\ & \Rightarrow {}^{3}{{C}_{2}}=\dfrac{3}{1}=3 \\\ \end{aligned}$
Substituting the above values of ${}^{4}{{C}_{2}}\And {}^{3}{{C}_{2}}$ in ${}^{4}{{C}_{2}}\times {}^{3}{{C}_{2}}$ we get,
${}^{4}{{C}_{2}}\times {}^{3}{{C}_{2}}=18$
From the above solution, we have found that 18 numbers of parallelograms can be formed from a set of four parallel lines intersecting three parallel lines.
Hence, the correct option is (b).

Note: The plausible mistake could be you wrongly write the expansion of combination factorials.
The expansion of ${}^{n}{{C}_{r}}$ is:
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
In the above formula, you might have mistakenly written n + r instead of n – r so be careful while writing the formulae.