There are $n$ points in a plane, no three of which collinear. Find Number of diagonals in a polygon of $n$ sides.
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Q-There are $n$ points in a plane, no three of which collinear. Find the number of diagonals in a polygon of $n$ sides.
Note I found error was with formula used to find number of triangles.
My attempt
In $n$ points there are $nchoose3$ triangles.
So, a polygon formed by $nchoose3$ triangles will have $nchoose3 + 2$ sides (I got this result by noticing series like square have 2 triangles, pentagon have 3 triangles,.....)
And a polygon of $n$ sides have $fracn(n-3)2$ diagonals .
Hence $nchoose3 + 2$ sides will have $frac12 Big[big( nchoose3big[nchoose3 + 2big] + 2big) -3Big]$ diagonals .
But in textbook it's answer is $nchoose2-n$
What mistake I had done. Please help me to find out error
Please note that I am a highschool student seeking help from teachers, so please don't close my question.
Thanks everyone for paying attention to my question!!!!!
combinatorics combinations
edited Aug 21 at 7:19
N. F. Taussig
38.6k93053
asked Aug 19 at 10:47
Rafael Nadal
1369
 |Â
show 3 more comments
up vote
1
down vote
favorite
Q-There are $n$ points in a plane, no three of which collinear. Find the number of diagonals in a polygon of $n$ sides.
Note I found error was with formula used to find number of triangles.
My attempt
In $n$ points there are $nchoose3$ triangles.
So, a polygon formed by $nchoose3$ triangles will have $nchoose3 + 2$ sides (I got this result by noticing series like square have 2 triangles, pentagon have 3 triangles,.....)
And a polygon of $n$ sides have $fracn(n-3)2$ diagonals .
Hence $nchoose3 + 2$ sides will have $frac12 Big[big( nchoose3big[nchoose3 + 2big] + 2big) -3Big]$ diagonals .
But in textbook it's answer is $nchoose2-n$
What mistake I had done. Please help me to find out error
Please note that I am a highschool student seeking help from teachers, so please don't close my question.
Thanks everyone for paying attention to my question!!!!!
combinatorics combinations
edited Aug 21 at 7:19
N. F. Taussig
38.6k93053
asked Aug 19 at 10:47
Rafael Nadal
1369
3
You construct diagonals with $2$ points... So, $binomn2$ . Again , the sides of polygon aren't diagonals... So, $binomn2-n$ .
– Entrepreneur
Aug 19 at 10:49
@Entrepreneur Thanks sir, I got it .But i want to know why answer is not coming by this method
– Rafael Nadal
Aug 19 at 10:54
How many triangles does a hexagon have? You must go through your first argument.
– Entrepreneur
Aug 19 at 10:58
@Entrepreneur Hexagon = 4 triangles it goes perfect with my pattern of n-2
– Rafael Nadal
Aug 19 at 11:01
No. It has $6$ of 'em
– Entrepreneur
Aug 19 at 11:05
 |Â
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Q-There are $n$ points in a plane, no three of which collinear. Find the number of diagonals in a polygon of $n$ sides.
Note I found error was with formula used to find number of triangles.
My attempt
In $n$ points there are $nchoose3$ triangles.
So, a polygon formed by $nchoose3$ triangles will have $nchoose3 + 2$ sides (I got this result by noticing series like square have 2 triangles, pentagon have 3 triangles,.....)
And a polygon of $n$ sides have $fracn(n-3)2$ diagonals .
Hence $nchoose3 + 2$ sides will have $frac12 Big[big( nchoose3big[nchoose3 + 2big] + 2big) -3Big]$ diagonals .
But in textbook it's answer is $nchoose2-n$
What mistake I had done. Please help me to find out error
Please note that I am a highschool student seeking help from teachers, so please don't close my question.
Thanks everyone for paying attention to my question!!!!!
combinatorics combinations
edited Aug 21 at 7:19
N. F. Taussig
38.6k93053
asked Aug 19 at 10:47
Rafael Nadal
1369
Q-There are $n$ points in a plane, no three of which collinear. Find the number of diagonals in a polygon of $n$ sides.
Note I found error was with formula used to find number of triangles.
My attempt
In $n$ points there are $nchoose3$ triangles.
So, a polygon formed by $nchoose3$ triangles will have $nchoose3 + 2$ sides (I got this result by noticing series like square have 2 triangles, pentagon have 3 triangles,.....)
And a polygon of $n$ sides have $fracn(n-3)2$ diagonals .
Hence $nchoose3 + 2$ sides will have $frac12 Big[big( nchoose3big[nchoose3 + 2big] + 2big) -3Big]$ diagonals .
But in textbook it's answer is $nchoose2-n$
What mistake I had done. Please help me to find out error
Please note that I am a highschool student seeking help from teachers, so please don't close my question.
Thanks everyone for paying attention to my question!!!!!
combinatorics combinations
edited Aug 21 at 7:19
N. F. Taussig
38.6k93053
asked Aug 19 at 10:47
Rafael Nadal
1369
edited Aug 21 at 7:19
N. F. Taussig
38.6k93053
edited Aug 21 at 7:19
N. F. Taussig
38.6k93053
edited Aug 21 at 7:19
N. F. Taussig
38.6k93053
38.6k93053
asked Aug 19 at 10:47
Rafael Nadal
1369
asked Aug 19 at 10:47
Rafael Nadal
1369
asked Aug 19 at 10:47
Rafael Nadal
1369
1369
3
You construct diagonals with $2$ points... So, $binomn2$ . Again , the sides of polygon aren't diagonals... So, $binomn2-n$ .
– Entrepreneur
Aug 19 at 10:49
@Entrepreneur Thanks sir, I got it .But i want to know why answer is not coming by this method
– Rafael Nadal
Aug 19 at 10:54
How many triangles does a hexagon have? You must go through your first argument.
– Entrepreneur
Aug 19 at 10:58
@Entrepreneur Hexagon = 4 triangles it goes perfect with my pattern of n-2
– Rafael Nadal
Aug 19 at 11:01
No. It has $6$ of 'em
– Entrepreneur
Aug 19 at 11:05
 |Â
show 3 more comments
3
You construct diagonals with $2$ points... So, $binomn2$ . Again , the sides of polygon aren't diagonals... So, $binomn2-n$ .
– Entrepreneur
Aug 19 at 10:49
@Entrepreneur Thanks sir, I got it .But i want to know why answer is not coming by this method
– Rafael Nadal
Aug 19 at 10:54
How many triangles does a hexagon have? You must go through your first argument.
– Entrepreneur
Aug 19 at 10:58
@Entrepreneur Hexagon = 4 triangles it goes perfect with my pattern of n-2
– Rafael Nadal
Aug 19 at 11:01
No. It has $6$ of 'em
– Entrepreneur
Aug 19 at 11:05
3
3
You construct diagonals with $2$ points... So, $binomn2$ . Again , the sides of polygon aren't diagonals... So, $binomn2-n$ .
– Entrepreneur
Aug 19 at 10:49
You construct diagonals with $2$ points... So, $binomn2$ . Again , the sides of polygon aren't diagonals... So, $binomn2-n$ .
– Entrepreneur
Aug 19 at 10:49
@Entrepreneur Thanks sir, I got it .But i want to know why answer is not coming by this method
– Rafael Nadal
Aug 19 at 10:54
@Entrepreneur Thanks sir, I got it .But i want to know why answer is not coming by this method
– Rafael Nadal
Aug 19 at 10:54
How many triangles does a hexagon have? You must go through your first argument.
– Entrepreneur
Aug 19 at 10:58
How many triangles does a hexagon have? You must go through your first argument.
– Entrepreneur
Aug 19 at 10:58
@Entrepreneur Hexagon = 4 triangles it goes perfect with my pattern of n-2
– Rafael Nadal
Aug 19 at 11:01
@Entrepreneur Hexagon = 4 triangles it goes perfect with my pattern of n-2
– Rafael Nadal
Aug 19 at 11:01
No. It has $6$ of 'em
– Entrepreneur
Aug 19 at 11:05
No. It has $6$ of 'em
– Entrepreneur
Aug 19 at 11:05
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
If you insist to make a conclusion starting from the number of triangles...
Say that you have $n$ points: $A_1, A_2, ..., A_n$.
These points define $binom n3$ triangles. The total number of edges in all these triangles is: $3timesbinom n3$. However each edge is counted several times. For example, edge $A_1A_2$ appears in triangles $A_1A_2A_3$, $A_1A_2A_4$, ..., $A_1A_2A_n$ which means that each edge appears exactly $n-2$ times in all possible triangles.
So the total number of lines is actually:
$$frac3times binom n3n-2$$
Not all lines are diagonals. You get the number of diagonas by subtracting the number of polygon sides $n$ from the total count of lines:
$$frac3times binom n3n-2-n=frac3timesfracn(n-1)(n-2)3times2times1n-2-n=fracn(n-1)2-n=binom n2-n$$
The most complicated way to count the number of diagonals, but you asked for it :)
answered Aug 20 at 13:09
Oldboy
2,8671318
There is a quicker way of getting to $n choose 2$ as the number of lines joining two of the $n$ points
– Henry
Aug 20 at 13:14
1
@Henry please read my introduction... and conclusion. Submitter asked for a triangle-based proof, not the easiest one. Each vertex of the polygon participates in exactly $n-3$ diagonals with each diagonal counted twice. So the total number of diagonals is $fracn(n-3)2$
– Oldboy
Aug 20 at 13:15
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
If you insist to make a conclusion starting from the number of triangles...
Say that you have $n$ points: $A_1, A_2, ..., A_n$.
These points define $binom n3$ triangles. The total number of edges in all these triangles is: $3timesbinom n3$. However each edge is counted several times. For example, edge $A_1A_2$ appears in triangles $A_1A_2A_3$, $A_1A_2A_4$, ..., $A_1A_2A_n$ which means that each edge appears exactly $n-2$ times in all possible triangles.
So the total number of lines is actually:
$$frac3times binom n3n-2$$
Not all lines are diagonals. You get the number of diagonas by subtracting the number of polygon sides $n$ from the total count of lines:
$$frac3times binom n3n-2-n=frac3timesfracn(n-1)(n-2)3times2times1n-2-n=fracn(n-1)2-n=binom n2-n$$
The most complicated way to count the number of diagonals, but you asked for it :)
answered Aug 20 at 13:09
Oldboy
2,8671318
There is a quicker way of getting to $n choose 2$ as the number of lines joining two of the $n$ points
– Henry
Aug 20 at 13:14
1
@Henry please read my introduction... and conclusion. Submitter asked for a triangle-based proof, not the easiest one. Each vertex of the polygon participates in exactly $n-3$ diagonals with each diagonal counted twice. So the total number of diagonals is $fracn(n-3)2$
– Oldboy
Aug 20 at 13:15
add a comment |Â
up vote
3
down vote
accepted
If you insist to make a conclusion starting from the number of triangles...
Say that you have $n$ points: $A_1, A_2, ..., A_n$.
These points define $binom n3$ triangles. The total number of edges in all these triangles is: $3timesbinom n3$. However each edge is counted several times. For example, edge $A_1A_2$ appears in triangles $A_1A_2A_3$, $A_1A_2A_4$, ..., $A_1A_2A_n$ which means that each edge appears exactly $n-2$ times in all possible triangles.
So the total number of lines is actually:
$$frac3times binom n3n-2$$
Not all lines are diagonals. You get the number of diagonas by subtracting the number of polygon sides $n$ from the total count of lines:
$$frac3times binom n3n-2-n=frac3timesfracn(n-1)(n-2)3times2times1n-2-n=fracn(n-1)2-n=binom n2-n$$
The most complicated way to count the number of diagonals, but you asked for it :)
answered Aug 20 at 13:09
Oldboy
2,8671318
There is a quicker way of getting to $n choose 2$ as the number of lines joining two of the $n$ points
– Henry
Aug 20 at 13:14
1
@Henry please read my introduction... and conclusion. Submitter asked for a triangle-based proof, not the easiest one. Each vertex of the polygon participates in exactly $n-3$ diagonals with each diagonal counted twice. So the total number of diagonals is $fracn(n-3)2$
– Oldboy
Aug 20 at 13:15
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
If you insist to make a conclusion starting from the number of triangles...
Say that you have $n$ points: $A_1, A_2, ..., A_n$.
These points define $binom n3$ triangles. The total number of edges in all these triangles is: $3timesbinom n3$. However each edge is counted several times. For example, edge $A_1A_2$ appears in triangles $A_1A_2A_3$, $A_1A_2A_4$, ..., $A_1A_2A_n$ which means that each edge appears exactly $n-2$ times in all possible triangles.
So the total number of lines is actually:
$$frac3times binom n3n-2$$
Not all lines are diagonals. You get the number of diagonas by subtracting the number of polygon sides $n$ from the total count of lines:
$$frac3times binom n3n-2-n=frac3timesfracn(n-1)(n-2)3times2times1n-2-n=fracn(n-1)2-n=binom n2-n$$
The most complicated way to count the number of diagonals, but you asked for it :)
answered Aug 20 at 13:09
Oldboy
2,8671318
If you insist to make a conclusion starting from the number of triangles...
Say that you have $n$ points: $A_1, A_2, ..., A_n$.
These points define $binom n3$ triangles. The total number of edges in all these triangles is: $3timesbinom n3$. However each edge is counted several times. For example, edge $A_1A_2$ appears in triangles $A_1A_2A_3$, $A_1A_2A_4$, ..., $A_1A_2A_n$ which means that each edge appears exactly $n-2$ times in all possible triangles.
So the total number of lines is actually:
$$frac3times binom n3n-2$$
Not all lines are diagonals. You get the number of diagonas by subtracting the number of polygon sides $n$ from the total count of lines:
$$frac3times binom n3n-2-n=frac3timesfracn(n-1)(n-2)3times2times1n-2-n=fracn(n-1)2-n=binom n2-n$$
The most complicated way to count the number of diagonals, but you asked for it :)
answered Aug 20 at 13:09
Oldboy
2,8671318
answered Aug 20 at 13:09
Oldboy
2,8671318
answered Aug 20 at 13:09
Oldboy
2,8671318
answered Aug 20 at 13:09
Oldboy
2,8671318
2,8671318
There is a quicker way of getting to $n choose 2$ as the number of lines joining two of the $n$ points
– Henry
Aug 20 at 13:14
1
@Henry please read my introduction... and conclusion. Submitter asked for a triangle-based proof, not the easiest one. Each vertex of the polygon participates in exactly $n-3$ diagonals with each diagonal counted twice. So the total number of diagonals is $fracn(n-3)2$
– Oldboy
Aug 20 at 13:15
add a comment |Â
There is a quicker way of getting to $n choose 2$ as the number of lines joining two of the $n$ points
– Henry
Aug 20 at 13:14
1
@Henry please read my introduction... and conclusion. Submitter asked for a triangle-based proof, not the easiest one. Each vertex of the polygon participates in exactly $n-3$ diagonals with each diagonal counted twice. So the total number of diagonals is $fracn(n-3)2$
– Oldboy
Aug 20 at 13:15
There is a quicker way of getting to $n choose 2$ as the number of lines joining two of the $n$ points
– Henry
Aug 20 at 13:14
There is a quicker way of getting to $n choose 2$ as the number of lines joining two of the $n$ points
– Henry
Aug 20 at 13:14
1
1
@Henry please read my introduction... and conclusion. Submitter asked for a triangle-based proof, not the easiest one. Each vertex of the polygon participates in exactly $n-3$ diagonals with each diagonal counted twice. So the total number of diagonals is $fracn(n-3)2$
– Oldboy
Aug 20 at 13:15
@Henry please read my introduction... and conclusion. Submitter asked for a triangle-based proof, not the easiest one. Each vertex of the polygon participates in exactly $n-3$ diagonals with each diagonal counted twice. So the total number of diagonals is $fracn(n-3)2$
– Oldboy
Aug 20 at 13:15
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2887582%2fthere-are-n-points-in-a-plane-no-three-of-which-collinear-find-number-of-dia%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
3
You construct diagonals with $2$ points... So, $binomn2$ . Again , the sides of polygon aren't diagonals... So, $binomn2-n$ .
– Entrepreneur
Aug 19 at 10:49
@Entrepreneur Thanks sir, I got it .But i want to know why answer is not coming by this method
– Rafael Nadal
Aug 19 at 10:54
How many triangles does a hexagon have? You must go through your first argument.
– Entrepreneur
Aug 19 at 10:58
@Entrepreneur Hexagon = 4 triangles it goes perfect with my pattern of n-2
– Rafael Nadal
Aug 19 at 11:01
No. It has $6$ of 'em
– Entrepreneur
Aug 19 at 11:05