Any 2 points determine a line. If there are 6 points in a plane, no 3 of which lie on the same line, how many lines are determined by pairs of these 6 points?
The answer key says 15, and the explanation is (6!)/(2!4!). Could someone explain to me why there is a 6, 2, and 4 factorials and how to approach this question please?
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You will find couple explanations.
To solve this problem, we need to consider the number of ways we can choose two points from the given six points to determine a line.
First, let's understand the factors involved in the calculation:
- The 6! (6 factorial) in the numerator represents the total number of ways we can arrange all six points.
- The 2! (2 factorial) in the denominator accounts for the fact that the order in which we select the two points does not matter. For example, choosing points A and B is the same as choosing points B and A.
- The 4! (4 factorial) in the denominator accounts for the fact that the remaining four points not chosen to determine each line can be arranged in a different order. This is necessary because we are only interested in unique lines.
Using these factors, we can calculate the total number of lines determined by pairs of these six points by dividing the total number of arrangements by the number of duplicate arrangements:
Number of lines = (6!) / (2! * 4!)
= (6 * 5 * 4 * 3 * 2 * 1) / (2 * 1 * 4 * 3 * 2 * 1)
= (6 * 5) / (2 * 1)
= 30 / 2
= 15
Therefore, there are 15 lines that can be determined by pairs of these six points.
To understand why there are factorials involved in this problem, let's break it down step by step:
We are given that there are 6 points in a plane, no 3 of which lie on the same line.
First, let's determine how many possible pairs of points we can make out of these 6 points.
To choose a pair of points, we need to choose 2 points out of the 6 available points. This is commonly represented as selecting 2 points "out of" 6, which can be written as "6 choose 2" or denoted as 6C2.
The formula for "n choose r" or "nCr" is:
nCr = n! / (r!(n - r)!)
where n! denotes the factorial of n (which is the product of all positive integers less than or equal to n), r! denotes the factorial of r, and (n - r)! denotes the factorial of (n - r).
In our case, we need to calculate 6C2, which is equivalent to:
6C2 = 6! / (2!(6-2)!)
Let's break down the equation:
6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
2! = 2 x 1 = 2
(6 - 2)! = 4! = 4 x 3 x 2 x 1 = 24
Plugging these values into the equation:
6C2 = 720 / (2 x 24)
Simplifying:
6C2 = 720 / 48
6C2 = 15
So, we have determined that there are 15 possible pairs of points we can form out of the 6 given points.
Now, let's determine how many lines are determined by these pairs of points.
For any two distinct points, there is exactly one line passing through them. So, each pair of points corresponds to a unique line.
Since we found that there are 15 possible pairs of points, there are also 15 lines determined by these pairs of points.
Therefore, the answer is 15.
This type of question often comes up when you study
"permutations" and "combinations"
You clearly must be studying permutations and combinations
In this case you are "choosing" any 2 of the 6 points
choose ----> combinations, (the order in which you choose any 2 points does not matter)
e.g. AB or BA yields the same line
the definition of C(n,r) is n!/(r!(n-r)!) , see your text
and we have to choose 2 from 6
= C(6,2) = 6!/(2!4!) = 15