No. of polygons in a polygon with no side coinciding.
$begingroup$
Here is the full question:-
r-sided polygons are formed by joining the vertices of an n-sided
polygon. Find the number of polygons that can be formed, none of whose
sides coincide with those of the n-sided polygon.
I imagined $(n-r)$ vertices in a closed polygon. There are $(n-r)$ possibilities for adding r vertices between them. (If we add r vertices here then no 2 vertices will be together). This leads me to $binom{n-r}{r}$. But the correct answer wants me to multiply it with $frac{n}{n-r}$
What is the need for the last step?
combinatorics geometry combinations
edited Dec 25 '18 at 2:56
Namaste
1
asked Dec 24 '18 at 14:40
Abhinay PandeyAbhinay Pandey
63
$endgroup$
add a comment |
$begingroup$
Here is the full question:-
r-sided polygons are formed by joining the vertices of an n-sided
polygon. Find the number of polygons that can be formed, none of whose
sides coincide with those of the n-sided polygon.
I imagined $(n-r)$ vertices in a closed polygon. There are $(n-r)$ possibilities for adding r vertices between them. (If we add r vertices here then no 2 vertices will be together). This leads me to $binom{n-r}{r}$. But the correct answer wants me to multiply it with $frac{n}{n-r}$
What is the need for the last step?
combinatorics geometry combinations
edited Dec 25 '18 at 2:56
Namaste
1
asked Dec 24 '18 at 14:40
Abhinay PandeyAbhinay Pandey
63
$endgroup$
$begingroup$
You can just check that it is correct for small numbers. For $n=6,r=3$ there are two choices instead of $1$. For $n=7, r=3$ there are seven instead of four as there is one case where two vertices are three apart and the first of those can be any of the seven vertices.
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:42
$begingroup$
There is a good discussion here as well. It is for $r=7$, but really applies more broadly.
$endgroup$
– Ross Millikan
Dec 25 '18 at 3:07
add a comment |
$begingroup$
Here is the full question:-
r-sided polygons are formed by joining the vertices of an n-sided
polygon. Find the number of polygons that can be formed, none of whose
sides coincide with those of the n-sided polygon.
I imagined $(n-r)$ vertices in a closed polygon. There are $(n-r)$ possibilities for adding r vertices between them. (If we add r vertices here then no 2 vertices will be together). This leads me to $binom{n-r}{r}$. But the correct answer wants me to multiply it with $frac{n}{n-r}$
What is the need for the last step?
combinatorics geometry combinations
edited Dec 25 '18 at 2:56
Namaste
1
asked Dec 24 '18 at 14:40
Abhinay PandeyAbhinay Pandey
63
$endgroup$
Here is the full question:-
r-sided polygons are formed by joining the vertices of an n-sided
polygon. Find the number of polygons that can be formed, none of whose
sides coincide with those of the n-sided polygon.
I imagined $(n-r)$ vertices in a closed polygon. There are $(n-r)$ possibilities for adding r vertices between them. (If we add r vertices here then no 2 vertices will be together). This leads me to $binom{n-r}{r}$. But the correct answer wants me to multiply it with $frac{n}{n-r}$
What is the need for the last step?
combinatorics geometry combinations
combinatorics geometry combinations
edited Dec 25 '18 at 2:56
Namaste
1
asked Dec 24 '18 at 14:40
Abhinay PandeyAbhinay Pandey
63
edited Dec 25 '18 at 2:56
Namaste
1
asked Dec 24 '18 at 14:40
Abhinay PandeyAbhinay Pandey
63
edited Dec 25 '18 at 2:56
Namaste
1
edited Dec 25 '18 at 2:56
Namaste
1
edited Dec 25 '18 at 2:56
Namaste
1
1
asked Dec 24 '18 at 14:40
Abhinay PandeyAbhinay Pandey
63
asked Dec 24 '18 at 14:40
Abhinay PandeyAbhinay Pandey
63
asked Dec 24 '18 at 14:40
Abhinay PandeyAbhinay Pandey
63
63
$begingroup$
You can just check that it is correct for small numbers. For $n=6,r=3$ there are two choices instead of $1$. For $n=7, r=3$ there are seven instead of four as there is one case where two vertices are three apart and the first of those can be any of the seven vertices.
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:42
$begingroup$
There is a good discussion here as well. It is for $r=7$, but really applies more broadly.
$endgroup$
– Ross Millikan
Dec 25 '18 at 3:07
add a comment |
$begingroup$
You can just check that it is correct for small numbers. For $n=6,r=3$ there are two choices instead of $1$. For $n=7, r=3$ there are seven instead of four as there is one case where two vertices are three apart and the first of those can be any of the seven vertices.
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:42
$begingroup$
There is a good discussion here as well. It is for $r=7$, but really applies more broadly.
$endgroup$
– Ross Millikan
Dec 25 '18 at 3:07
$begingroup$
You can just check that it is correct for small numbers. For $n=6,r=3$ there are two choices instead of $1$. For $n=7, r=3$ there are seven instead of four as there is one case where two vertices are three apart and the first of those can be any of the seven vertices.
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:42
$begingroup$
You can just check that it is correct for small numbers. For $n=6,r=3$ there are two choices instead of $1$. For $n=7, r=3$ there are seven instead of four as there is one case where two vertices are three apart and the first of those can be any of the seven vertices.
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:42
$begingroup$
There is a good discussion here as well. It is for $r=7$, but really applies more broadly.
$endgroup$
– Ross Millikan
Dec 25 '18 at 3:07
$begingroup$
There is a good discussion here as well. It is for $r=7$, but really applies more broadly.
$endgroup$
– Ross Millikan
Dec 25 '18 at 3:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Look at the case $n=6,r=3$. You have a hexagon with vertices numbered $1$ to $6$, and there are two triangles you can make in this hexagon, with vertices numbered $1,3,5$ and $2,4,6$. But your formula only counts one of these.
Look at your method. You start with $n-r=3$ vertices, which are distinct. Say they are numbered $1,2,3$. Then you select $r=3$ of these vertices, and insert a vertex next to them. This results in
$$
1_2_3_
$$
Now you have to choose the labels for those inserted vertices. This part you have not accounted for. In the final result, the vertices need to be numbered $1$ to $6$ in order, so one way to do this is just to start at $1$, and rename the vertices $2$ through $6$ in order, obtaining
$$
1underline23underline45underline6
$$
This gives the triangle $135$.
This illustrates the following problem with your method; $binom{n-r}r$ counts the number of ways to choose a polygon where vertex number $1$ is included. Therefore we need to multiply by $n$, to also include the number of polygons which use vertices $2,3dots,n$. However, this will over-count the polygons by a factor of $n-r$, so you must divide by that in the end.
answered Dec 24 '18 at 16:35
Mike EarnestMike Earnest
24.3k22151
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Look at the case $n=6,r=3$. You have a hexagon with vertices numbered $1$ to $6$, and there are two triangles you can make in this hexagon, with vertices numbered $1,3,5$ and $2,4,6$. But your formula only counts one of these.
Look at your method. You start with $n-r=3$ vertices, which are distinct. Say they are numbered $1,2,3$. Then you select $r=3$ of these vertices, and insert a vertex next to them. This results in
$$
1_2_3_
$$
Now you have to choose the labels for those inserted vertices. This part you have not accounted for. In the final result, the vertices need to be numbered $1$ to $6$ in order, so one way to do this is just to start at $1$, and rename the vertices $2$ through $6$ in order, obtaining
$$
1underline23underline45underline6
$$
This gives the triangle $135$.
This illustrates the following problem with your method; $binom{n-r}r$ counts the number of ways to choose a polygon where vertex number $1$ is included. Therefore we need to multiply by $n$, to also include the number of polygons which use vertices $2,3dots,n$. However, this will over-count the polygons by a factor of $n-r$, so you must divide by that in the end.
answered Dec 24 '18 at 16:35
Mike EarnestMike Earnest
24.3k22151
$endgroup$
add a comment |
$begingroup$
Look at the case $n=6,r=3$. You have a hexagon with vertices numbered $1$ to $6$, and there are two triangles you can make in this hexagon, with vertices numbered $1,3,5$ and $2,4,6$. But your formula only counts one of these.
Look at your method. You start with $n-r=3$ vertices, which are distinct. Say they are numbered $1,2,3$. Then you select $r=3$ of these vertices, and insert a vertex next to them. This results in
$$
1_2_3_
$$
Now you have to choose the labels for those inserted vertices. This part you have not accounted for. In the final result, the vertices need to be numbered $1$ to $6$ in order, so one way to do this is just to start at $1$, and rename the vertices $2$ through $6$ in order, obtaining
$$
1underline23underline45underline6
$$
This gives the triangle $135$.
This illustrates the following problem with your method; $binom{n-r}r$ counts the number of ways to choose a polygon where vertex number $1$ is included. Therefore we need to multiply by $n$, to also include the number of polygons which use vertices $2,3dots,n$. However, this will over-count the polygons by a factor of $n-r$, so you must divide by that in the end.
answered Dec 24 '18 at 16:35
Mike EarnestMike Earnest
24.3k22151
$endgroup$
add a comment |
$begingroup$
Look at the case $n=6,r=3$. You have a hexagon with vertices numbered $1$ to $6$, and there are two triangles you can make in this hexagon, with vertices numbered $1,3,5$ and $2,4,6$. But your formula only counts one of these.
Look at your method. You start with $n-r=3$ vertices, which are distinct. Say they are numbered $1,2,3$. Then you select $r=3$ of these vertices, and insert a vertex next to them. This results in
$$
1_2_3_
$$
Now you have to choose the labels for those inserted vertices. This part you have not accounted for. In the final result, the vertices need to be numbered $1$ to $6$ in order, so one way to do this is just to start at $1$, and rename the vertices $2$ through $6$ in order, obtaining
$$
1underline23underline45underline6
$$
This gives the triangle $135$.
This illustrates the following problem with your method; $binom{n-r}r$ counts the number of ways to choose a polygon where vertex number $1$ is included. Therefore we need to multiply by $n$, to also include the number of polygons which use vertices $2,3dots,n$. However, this will over-count the polygons by a factor of $n-r$, so you must divide by that in the end.
answered Dec 24 '18 at 16:35
Mike EarnestMike Earnest
24.3k22151
$endgroup$
Look at the case $n=6,r=3$. You have a hexagon with vertices numbered $1$ to $6$, and there are two triangles you can make in this hexagon, with vertices numbered $1,3,5$ and $2,4,6$. But your formula only counts one of these.
Look at your method. You start with $n-r=3$ vertices, which are distinct. Say they are numbered $1,2,3$. Then you select $r=3$ of these vertices, and insert a vertex next to them. This results in
$$
1_2_3_
$$
Now you have to choose the labels for those inserted vertices. This part you have not accounted for. In the final result, the vertices need to be numbered $1$ to $6$ in order, so one way to do this is just to start at $1$, and rename the vertices $2$ through $6$ in order, obtaining
$$
1underline23underline45underline6
$$
This gives the triangle $135$.
This illustrates the following problem with your method; $binom{n-r}r$ counts the number of ways to choose a polygon where vertex number $1$ is included. Therefore we need to multiply by $n$, to also include the number of polygons which use vertices $2,3dots,n$. However, this will over-count the polygons by a factor of $n-r$, so you must divide by that in the end.
answered Dec 24 '18 at 16:35
Mike EarnestMike Earnest
24.3k22151
answered Dec 24 '18 at 16:35
Mike EarnestMike Earnest
24.3k22151
answered Dec 24 '18 at 16:35
Mike EarnestMike Earnest
24.3k22151
answered Dec 24 '18 at 16:35
Mike EarnestMike Earnest
24.3k22151
24.3k22151
add a comment |
add a comment |
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$begingroup$
You can just check that it is correct for small numbers. For $n=6,r=3$ there are two choices instead of $1$. For $n=7, r=3$ there are seven instead of four as there is one case where two vertices are three apart and the first of those can be any of the seven vertices.
$endgroup$
– Ross Millikan
Dec 24 '18 at 15:42
$begingroup$
There is a good discussion here as well. It is for $r=7$, but really applies more broadly.
$endgroup$
– Ross Millikan
Dec 25 '18 at 3:07