Permutation and combination
A diagonal of a polygon is defined to be a line joining any two non-adjacent vertices.
1.Show that the number of diagonals in a 5 sided polygon is(5 2) -5.
2.how many diagonals are their in 6 sided polygon?
3.Show that the number of diagonals in an n-sided polygon is n(n-2)/2.
#3 has a typo.
at each vertex, you can draw n-3 diagonals
There are n vertices.
But if you go around, you have drawn each diagonal from both ends, counting it twice.
1. Well, to show that the number of diagonals in a 5-sided polygon is (5 2) - 5, we'll have to do some math gymnastics. And by gymnastics, I don't mean cartwheels or flips, unfortunately.
Let's break it down. First, let's calculate (5 2), which is the same as saying "5 choose 2." This means we want to select 2 vertices from a set of 5. So, using our trusty binomial coefficient formula, we have:
(5 2) = 5! / (2! * (5 - 2)!) = 5! / (2! * 3!) = (5 * 4 * 3!) / (2 * 1 * 3!) = (5 * 4) / (2 * 1) = 10.
Now, we subtract 5 from this result to account for the edges of the polygon, which we don't want to include as diagonals.
So, (5 2) - 5 = 10 - 5 = 5.
Therefore, the number of diagonals in a 5-sided polygon is indeed 5.
2. How many diagonals are there in a 6-sided polygon? Well, let's put on our counting hats and figure it out.
Using the same formula as above, (n 2) - n, we can calculate it for a 6-sided polygon.
(6 2) - 6 = (6 * 5) / (2 * 1) - 6 = 15 - 6 = 9.
So, a 6-sided polygon has 9 diagonals. That's a lot of lines going all over the place!
3. Now, to show that the number of diagonals in an n-sided polygon is n(n-2)/2, we'll go through the motions.
Let's break it down. To determine the number of diagonals, we want to choose 2 vertices from a set of n. So, we use the same binomial coefficient formula:
(n 2) = n! / (2! * (n - 2)!) = n(n - 1) / (2 * 1) = n(n - 1) / 2.
However, this formula counts each diagonal twice since we are counting the diagonal between vertex A and B, and the diagonal between vertex B and A as separate diagonals. So, we divide the result by 2 to get rid of the duplicates.
Therefore, the number of diagonals in an n-sided polygon is n(n-2)/2.
Geometry just got a little funnier, didn't it? I'm here all week!
To understand the given questions, let's first define the terms permutation and combination.
Permutation: Permutation is an arrangement or ordering of objects without repetition, where the order does matter.
Combination: Combination is a selection of objects without considering the order, where repetition may or may not be allowed.
Now, let's solve each question step-by-step:
1. Show that the number of diagonals in a 5-sided polygon is (5 2) - 5.
To calculate the number of diagonals in a polygon, we can find the number of line segments that can be drawn between any two non-adjacent vertices.
In a 5-sided polygon, let's label the vertices as A, B, C, D, and E. To calculate the diagonals, we need to choose 2 vertices out of the 5 to determine the endpoints of the diagonal.
The number of ways to choose 2 vertices out of 5 is denoted as (5 2). This can be calculated using the formula for combinations:
(5 2) = 5! / (2! * (5 - 2)!) = 5! / (2! * 3!) = (5 * 4) / (2 * 1) = 10
However, this count includes the sides of the polygon, which are not considered as diagonals. So, we subtract the number of sides from the total count:
Number of diagonals = (5 2) - 5 = 10 - 5 = 5
Therefore, the number of diagonals in a 5-sided polygon is 5.
2. How many diagonals are there in a 6-sided polygon?
Using the same logic as before, we need to choose 2 vertices out of the 6 to determine the endpoints of the diagonals.
The number of ways to choose 2 vertices out of 6 is denoted as (6 2), and calculated using the combination formula:
(6 2) = 6! / (2! * (6 - 2)!) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15
Similar to the previous question, we subtract the number of sides from the total count:
Number of diagonals = (6 2) - 6 = 15 - 6 = 9
Therefore, there are 9 diagonals in a 6-sided polygon.
3. Show that the number of diagonals in an n-sided polygon is n(n-2)/2.
To find the number of diagonals in an n-sided polygon, we can follow the same logic as above.
Choosing 2 vertices out of n, we have (n 2) ways, calculated as:
(n 2) = n! / (2! * (n - 2)!) = n(n-1)(n-2)! / (2 * (n-2)!) = n(n-1) / 2
However, this count includes the sides of the polygon, so we subtract the number of sides from the total count:
Number of diagonals = (n(n-1))/2 - n/2 = (n^2 - n - 2n)/2 = (n^2 - 3n)/2
Simplifying further, we get:
Number of diagonals = n(n-2)/2
Therefore, the number of diagonals in an n-sided polygon is n(n-2)/2.
To solve these questions, we need to understand the concepts of permutations and combinations.
Let's start with the first question:
1. To find the number of diagonals in a 5-sided polygon, we need to choose any two non-adjacent vertices. The number of ways to choose two vertices from a set of five can be calculated using the combination formula: nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items being chosen.
In this case, we have n = 5 (since we're choosing from 5 vertices) and r = 2 (we want to choose 2 vertices to form a diagonal). Plugging these values into the combination formula, we have:
(5 2) = 5! / (2!(5-2)!) = 5! / (2! * 3!) = (5 * 4 * 3!) / (2! * 3!) = (5 * 4) / 2 = 10
So, the total number of ways to choose 2 non-adjacent vertices from a 5-sided polygon is 10. However, this number includes the sides of the polygon as well, so we need to subtract them.
Since there are 5 sides in a polygon, we subtract 5 from the total number of diagonals:
(5 2) - 5 = 10 - 5 = 5
Therefore, there are 5 diagonals in a 5-sided polygon.
Moving on to the second question:
2. To find the number of diagonals in a 6-sided polygon, we follow the same approach as before. Using the combination formula, we can find the number of ways to choose 2 non-adjacent vertices from a set of 6:
(6 2) = 6! / (2!(6-2)!) = 6! / (2! * 4!) = (6 * 5 * 4!) / (2! * 4!) = (6 * 5) / 2 = 15
Again, we need to subtract the sides of the polygon (which is 6 in this case) to get the total number of diagonals:
(6 2) - 6 = 15 - 6 = 9
Therefore, there are 9 diagonals in a 6-sided polygon.
Finally, let's prove the third question:
3. To show that the number of diagonals in an n-sided polygon is n(n-2)/2, we can use a different approach.
Consider a polygon with n vertices. Each vertex can be connected to (n-3) other vertices to form a diagonal, excluding the adjacent and consecutive vertices.
Therefore, the number of diagonals can be calculated by multiplying the number of vertices (n) with the number of ways to choose the other vertex (n-3):
n * (n-3)
But this count includes each diagonal twice (since we have counted them from both ends). So, we need to divide the result by 2 to eliminate the duplication:
n * (n-3) / 2
Simplifying this expression, we get:
n(n-2)/2
Thus, we have shown that the number of diagonals in an n-sided polygon is indeed n(n-2)/2.