- Md.Nazrul Islam
- Senior | Next Rank: 100 Posts
- Posts: 32
- Joined: Sat Jul 16, 2011 11:31 am
The number of parallelograms that can be formed from a set of four parallel straight line intersecting a set of three parallel straight lines .
Md.Nazrul Islam wrote:The number of parallelograms that can be formed from a set of four parallel straight line intersecting a set of three parallel straight lines .
From the set of 4 parallel lines:Md.Nazrul Islam wrote:The number of parallelograms that can be formed from a set of four parallel straight line intersecting a set of three parallel straight lines .
We use combinations and not permutations because we can`t change the order of the lines, right??!GMATGuruNY wrote:From the set of 4 parallel lines:Md.Nazrul Islam wrote:The number of parallelograms that can be formed from a set of four parallel straight line intersecting a set of three parallel straight lines .
We need to choose a combination of 2 to serve as 2 parallel sides of the parallelogram.
The number of combinations of 2 that can formed from 4 choices = 4C2 = 6.
From the set of 3 parallel lines:
We need to choose a combination of 2 to serve as the OTHER 2 parallel sides of the parallelogram.
The number of combinations of 2 that can be formed from 3 choices = 3C2 = 3.
To combine the options above, we multiply:
6*3 = 18.
We use combination because we need to chose.ka_t_rin wrote:We use combinations and not permutations because we can`t change the order of the lines, right??!GMATGuruNY wrote:From the set of 4 parallel lines:Md.Nazrul Islam wrote:The number of parallelograms that can be formed from a set of four parallel straight line intersecting a set of three parallel straight lines .
We need to choose a combination of 2 to serve as 2 parallel sides of the parallelogram.
The number of combinations of 2 that can formed from 4 choices = 4C2 = 6.
From the set of 3 parallel lines:
We need to choose a combination of 2 to serve as the OTHER 2 parallel sides of the parallelogram.
The number of combinations of 2 that can be formed from 3 choices = 3C2 = 3.
To combine the options above, we multiply:
6*3 = 18.