Exercise with Circle Permutations in Combinatorics
$begingroup$
I have a problem in solving an exercise. It says:
$10$ cars are about to park in a circle plaza, which $3$ of them are cars of permanent residents. In how many ways can the cars be put so as to:
a) None of the permanent residents' cars are parked in consecutive places with an another permanent resident's car ?
b) At least $2$ permanent residents' cars are parked in consecutive places ?
I translated the exercise from Greek so I apologise in advance if I have not phrased something correctly.
My thought :
a) From the total of 10 cars, 3 of them belong to permanent residents and the other 7 to "temporary" residents so I will refer to them from now on with $P$ and $T$ respectively.
It is obvious that we have a case of circular arrangement here ( I think that's how is called in English). We need a part of the elements of the circle to be always arranged in a specific way so I considered it a good idea to create a group of objects : $G = (PTPTP)$ which consists of the $3$ $P$ cars and any $2$ of the $7$ $T$ cars and with that if my thinking is correct we have to calculate the number of circles of 6 objects $(T,T,T,T,T,G)$ and multiply the result with $3!*frac{7!}{5!}$ which I think that they are the circles of the group. That gets me $5!*3!*6*7$ which is far away from the answer of the book which is $6!*frac{7!}{4!}$. Obviously I'm making something wrong.
b) I have no clue on how to solve b) any hint or help would be appreciated.
combinatorics
asked Dec 10 '18 at 15:02
NickDeltaNickDelta
285
$endgroup$
|
show 3 more comments
$begingroup$
I have a problem in solving an exercise. It says:
$10$ cars are about to park in a circle plaza, which $3$ of them are cars of permanent residents. In how many ways can the cars be put so as to:
a) None of the permanent residents' cars are parked in consecutive places with an another permanent resident's car ?
b) At least $2$ permanent residents' cars are parked in consecutive places ?
I translated the exercise from Greek so I apologise in advance if I have not phrased something correctly.
My thought :
a) From the total of 10 cars, 3 of them belong to permanent residents and the other 7 to "temporary" residents so I will refer to them from now on with $P$ and $T$ respectively.
It is obvious that we have a case of circular arrangement here ( I think that's how is called in English). We need a part of the elements of the circle to be always arranged in a specific way so I considered it a good idea to create a group of objects : $G = (PTPTP)$ which consists of the $3$ $P$ cars and any $2$ of the $7$ $T$ cars and with that if my thinking is correct we have to calculate the number of circles of 6 objects $(T,T,T,T,T,G)$ and multiply the result with $3!*frac{7!}{5!}$ which I think that they are the circles of the group. That gets me $5!*3!*6*7$ which is far away from the answer of the book which is $6!*frac{7!}{4!}$. Obviously I'm making something wrong.
b) I have no clue on how to solve b) any hint or help would be appreciated.
combinatorics
asked Dec 10 '18 at 15:02
NickDeltaNickDelta
285
$endgroup$
$begingroup$
Now, I believe that I know what you mean when you say "circle square", but keep in mind that this is a math forum. We will think of "square" as a rectangle with equal sides before we think of "square" as a plaza.
$endgroup$
– Arthur
Dec 10 '18 at 15:06
$begingroup$
Didn't have that thought when I was translating, I'm going to edit it and put plaza to avoid confusion.
$endgroup$
– NickDelta
Dec 10 '18 at 15:08
$begingroup$
To be clear, in our circular plaza, do we care which direction north is? I.e. if we label the cars $1,2,3,dots$ and we start listing which car parked where from the northernmost space going clockwise is the parking sequence $(1,2,3,dots,9,10)$ considered the same or different than the parking sequence $(2,3,4,dots,10,1)$?
$endgroup$
– JMoravitz
Dec 10 '18 at 15:14
$begingroup$
JMoravitz I think it's considered the same since we're talking for circle permutations and not simple ones
$endgroup$
– NickDelta
Dec 10 '18 at 15:17
1
$begingroup$
As for part (b), notice that the if it is not true that no permanent residents are adjacent then it must be true that at least two permanent residents are adjacent.
$endgroup$
– JMoravitz
Dec 10 '18 at 15:21
|
show 3 more comments
$begingroup$
I have a problem in solving an exercise. It says:
$10$ cars are about to park in a circle plaza, which $3$ of them are cars of permanent residents. In how many ways can the cars be put so as to:
a) None of the permanent residents' cars are parked in consecutive places with an another permanent resident's car ?
b) At least $2$ permanent residents' cars are parked in consecutive places ?
I translated the exercise from Greek so I apologise in advance if I have not phrased something correctly.
My thought :
a) From the total of 10 cars, 3 of them belong to permanent residents and the other 7 to "temporary" residents so I will refer to them from now on with $P$ and $T$ respectively.
It is obvious that we have a case of circular arrangement here ( I think that's how is called in English). We need a part of the elements of the circle to be always arranged in a specific way so I considered it a good idea to create a group of objects : $G = (PTPTP)$ which consists of the $3$ $P$ cars and any $2$ of the $7$ $T$ cars and with that if my thinking is correct we have to calculate the number of circles of 6 objects $(T,T,T,T,T,G)$ and multiply the result with $3!*frac{7!}{5!}$ which I think that they are the circles of the group. That gets me $5!*3!*6*7$ which is far away from the answer of the book which is $6!*frac{7!}{4!}$. Obviously I'm making something wrong.
b) I have no clue on how to solve b) any hint or help would be appreciated.
combinatorics
asked Dec 10 '18 at 15:02
NickDeltaNickDelta
285
$endgroup$
I have a problem in solving an exercise. It says:
$10$ cars are about to park in a circle plaza, which $3$ of them are cars of permanent residents. In how many ways can the cars be put so as to:
a) None of the permanent residents' cars are parked in consecutive places with an another permanent resident's car ?
b) At least $2$ permanent residents' cars are parked in consecutive places ?
I translated the exercise from Greek so I apologise in advance if I have not phrased something correctly.
My thought :
a) From the total of 10 cars, 3 of them belong to permanent residents and the other 7 to "temporary" residents so I will refer to them from now on with $P$ and $T$ respectively.
It is obvious that we have a case of circular arrangement here ( I think that's how is called in English). We need a part of the elements of the circle to be always arranged in a specific way so I considered it a good idea to create a group of objects : $G = (PTPTP)$ which consists of the $3$ $P$ cars and any $2$ of the $7$ $T$ cars and with that if my thinking is correct we have to calculate the number of circles of 6 objects $(T,T,T,T,T,G)$ and multiply the result with $3!*frac{7!}{5!}$ which I think that they are the circles of the group. That gets me $5!*3!*6*7$ which is far away from the answer of the book which is $6!*frac{7!}{4!}$. Obviously I'm making something wrong.
b) I have no clue on how to solve b) any hint or help would be appreciated.
combinatorics
combinatorics
asked Dec 10 '18 at 15:02
NickDeltaNickDelta
285
asked Dec 10 '18 at 15:02
NickDeltaNickDelta
285
edited Dec 10 '18 at 15:09
NickDelta
asked Dec 10 '18 at 15:02
NickDeltaNickDelta
285
asked Dec 10 '18 at 15:02
NickDeltaNickDelta
285
asked Dec 10 '18 at 15:02
NickDeltaNickDelta
285
285
$begingroup$
Now, I believe that I know what you mean when you say "circle square", but keep in mind that this is a math forum. We will think of "square" as a rectangle with equal sides before we think of "square" as a plaza.
$endgroup$
– Arthur
Dec 10 '18 at 15:06
$begingroup$
Didn't have that thought when I was translating, I'm going to edit it and put plaza to avoid confusion.
$endgroup$
– NickDelta
Dec 10 '18 at 15:08
$begingroup$
To be clear, in our circular plaza, do we care which direction north is? I.e. if we label the cars $1,2,3,dots$ and we start listing which car parked where from the northernmost space going clockwise is the parking sequence $(1,2,3,dots,9,10)$ considered the same or different than the parking sequence $(2,3,4,dots,10,1)$?
$endgroup$
– JMoravitz
Dec 10 '18 at 15:14
$begingroup$
JMoravitz I think it's considered the same since we're talking for circle permutations and not simple ones
$endgroup$
– NickDelta
Dec 10 '18 at 15:17
1
$begingroup$
As for part (b), notice that the if it is not true that no permanent residents are adjacent then it must be true that at least two permanent residents are adjacent.
$endgroup$
– JMoravitz
Dec 10 '18 at 15:21
|
show 3 more comments
$begingroup$
Now, I believe that I know what you mean when you say "circle square", but keep in mind that this is a math forum. We will think of "square" as a rectangle with equal sides before we think of "square" as a plaza.
$endgroup$
– Arthur
Dec 10 '18 at 15:06
$begingroup$
Didn't have that thought when I was translating, I'm going to edit it and put plaza to avoid confusion.
$endgroup$
– NickDelta
Dec 10 '18 at 15:08
$begingroup$
To be clear, in our circular plaza, do we care which direction north is? I.e. if we label the cars $1,2,3,dots$ and we start listing which car parked where from the northernmost space going clockwise is the parking sequence $(1,2,3,dots,9,10)$ considered the same or different than the parking sequence $(2,3,4,dots,10,1)$?
$endgroup$
– JMoravitz
Dec 10 '18 at 15:14
$begingroup$
JMoravitz I think it's considered the same since we're talking for circle permutations and not simple ones
$endgroup$
– NickDelta
Dec 10 '18 at 15:17
1
$begingroup$
As for part (b), notice that the if it is not true that no permanent residents are adjacent then it must be true that at least two permanent residents are adjacent.
$endgroup$
– JMoravitz
Dec 10 '18 at 15:21
$begingroup$
Now, I believe that I know what you mean when you say "circle square", but keep in mind that this is a math forum. We will think of "square" as a rectangle with equal sides before we think of "square" as a plaza.
$endgroup$
– Arthur
Dec 10 '18 at 15:06
$begingroup$
Now, I believe that I know what you mean when you say "circle square", but keep in mind that this is a math forum. We will think of "square" as a rectangle with equal sides before we think of "square" as a plaza.
$endgroup$
– Arthur
Dec 10 '18 at 15:06
$begingroup$
Didn't have that thought when I was translating, I'm going to edit it and put plaza to avoid confusion.
$endgroup$
– NickDelta
Dec 10 '18 at 15:08
$begingroup$
Didn't have that thought when I was translating, I'm going to edit it and put plaza to avoid confusion.
$endgroup$
– NickDelta
Dec 10 '18 at 15:08
$begingroup$
To be clear, in our circular plaza, do we care which direction north is? I.e. if we label the cars $1,2,3,dots$ and we start listing which car parked where from the northernmost space going clockwise is the parking sequence $(1,2,3,dots,9,10)$ considered the same or different than the parking sequence $(2,3,4,dots,10,1)$?
$endgroup$
– JMoravitz
Dec 10 '18 at 15:14
$begingroup$
To be clear, in our circular plaza, do we care which direction north is? I.e. if we label the cars $1,2,3,dots$ and we start listing which car parked where from the northernmost space going clockwise is the parking sequence $(1,2,3,dots,9,10)$ considered the same or different than the parking sequence $(2,3,4,dots,10,1)$?
$endgroup$
– JMoravitz
Dec 10 '18 at 15:14
$begingroup$
JMoravitz I think it's considered the same since we're talking for circle permutations and not simple ones
$endgroup$
– NickDelta
Dec 10 '18 at 15:17
$begingroup$
JMoravitz I think it's considered the same since we're talking for circle permutations and not simple ones
$endgroup$
– NickDelta
Dec 10 '18 at 15:17
1
1
$begingroup$
As for part (b), notice that the if it is not true that no permanent residents are adjacent then it must be true that at least two permanent residents are adjacent.
$endgroup$
– JMoravitz
Dec 10 '18 at 15:21
$begingroup$
As for part (b), notice that the if it is not true that no permanent residents are adjacent then it must be true that at least two permanent residents are adjacent.
$endgroup$
– JMoravitz
Dec 10 '18 at 15:21
|
show 3 more comments
1 Answer
1
active
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$begingroup$
For a, I don't understand your logic. You are missing configurations like $PTTPTTPTTT$. Let us attach a $T$ after each $P$. We can do that in $7cdot 6 cdot 5$ ways. We can then put our seven objects in order in $7!$ ways. We then divide by $7$ because each order can be rotated to start with any object. That gives $7!cdot 6 cdot 5$.
For b, we can just compute the total number of orders and subtract the answer to a. There are $9!$ orders that start with a given car, so the answer is $9!-7!cdot 6 cdot 5$
answered Dec 10 '18 at 15:19
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
You're right I was missing a lot of configurations with my thought. Your results are correct but sorry for my ignorance but I didn't understood your thought. I'm not a native English speaker and understanding maths in English is even more difficult than speaking it. Could you please be a little more analytic ?
$endgroup$
– NickDelta
Dec 10 '18 at 15:32
$begingroup$
What are you having trouble with? Attaching a T to each P seems to me like what you were doing and makes sure you don't have two Ps together.
$endgroup$
– Ross Millikan
Dec 10 '18 at 15:43
$begingroup$
Now i got it that phrase is equivalent to $frac{7!}(4!}$ . Very nice answer, thanks once again.
$endgroup$
– NickDelta
Dec 10 '18 at 15:47
add a comment |
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$begingroup$
For a, I don't understand your logic. You are missing configurations like $PTTPTTPTTT$. Let us attach a $T$ after each $P$. We can do that in $7cdot 6 cdot 5$ ways. We can then put our seven objects in order in $7!$ ways. We then divide by $7$ because each order can be rotated to start with any object. That gives $7!cdot 6 cdot 5$.
For b, we can just compute the total number of orders and subtract the answer to a. There are $9!$ orders that start with a given car, so the answer is $9!-7!cdot 6 cdot 5$
answered Dec 10 '18 at 15:19
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
You're right I was missing a lot of configurations with my thought. Your results are correct but sorry for my ignorance but I didn't understood your thought. I'm not a native English speaker and understanding maths in English is even more difficult than speaking it. Could you please be a little more analytic ?
$endgroup$
– NickDelta
Dec 10 '18 at 15:32
$begingroup$
What are you having trouble with? Attaching a T to each P seems to me like what you were doing and makes sure you don't have two Ps together.
$endgroup$
– Ross Millikan
Dec 10 '18 at 15:43
$begingroup$
Now i got it that phrase is equivalent to $frac{7!}(4!}$ . Very nice answer, thanks once again.
$endgroup$
– NickDelta
Dec 10 '18 at 15:47
add a comment |
$begingroup$
For a, I don't understand your logic. You are missing configurations like $PTTPTTPTTT$. Let us attach a $T$ after each $P$. We can do that in $7cdot 6 cdot 5$ ways. We can then put our seven objects in order in $7!$ ways. We then divide by $7$ because each order can be rotated to start with any object. That gives $7!cdot 6 cdot 5$.
For b, we can just compute the total number of orders and subtract the answer to a. There are $9!$ orders that start with a given car, so the answer is $9!-7!cdot 6 cdot 5$
answered Dec 10 '18 at 15:19
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
You're right I was missing a lot of configurations with my thought. Your results are correct but sorry for my ignorance but I didn't understood your thought. I'm not a native English speaker and understanding maths in English is even more difficult than speaking it. Could you please be a little more analytic ?
$endgroup$
– NickDelta
Dec 10 '18 at 15:32
$begingroup$
What are you having trouble with? Attaching a T to each P seems to me like what you were doing and makes sure you don't have two Ps together.
$endgroup$
– Ross Millikan
Dec 10 '18 at 15:43
$begingroup$
Now i got it that phrase is equivalent to $frac{7!}(4!}$ . Very nice answer, thanks once again.
$endgroup$
– NickDelta
Dec 10 '18 at 15:47
add a comment |
$begingroup$
For a, I don't understand your logic. You are missing configurations like $PTTPTTPTTT$. Let us attach a $T$ after each $P$. We can do that in $7cdot 6 cdot 5$ ways. We can then put our seven objects in order in $7!$ ways. We then divide by $7$ because each order can be rotated to start with any object. That gives $7!cdot 6 cdot 5$.
For b, we can just compute the total number of orders and subtract the answer to a. There are $9!$ orders that start with a given car, so the answer is $9!-7!cdot 6 cdot 5$
answered Dec 10 '18 at 15:19
Ross MillikanRoss Millikan
295k23198371
$endgroup$
For a, I don't understand your logic. You are missing configurations like $PTTPTTPTTT$. Let us attach a $T$ after each $P$. We can do that in $7cdot 6 cdot 5$ ways. We can then put our seven objects in order in $7!$ ways. We then divide by $7$ because each order can be rotated to start with any object. That gives $7!cdot 6 cdot 5$.
For b, we can just compute the total number of orders and subtract the answer to a. There are $9!$ orders that start with a given car, so the answer is $9!-7!cdot 6 cdot 5$
answered Dec 10 '18 at 15:19
Ross MillikanRoss Millikan
295k23198371
answered Dec 10 '18 at 15:19
Ross MillikanRoss Millikan
295k23198371
answered Dec 10 '18 at 15:19
Ross MillikanRoss Millikan
295k23198371
answered Dec 10 '18 at 15:19
Ross MillikanRoss Millikan
295k23198371
295k23198371
$begingroup$
You're right I was missing a lot of configurations with my thought. Your results are correct but sorry for my ignorance but I didn't understood your thought. I'm not a native English speaker and understanding maths in English is even more difficult than speaking it. Could you please be a little more analytic ?
$endgroup$
– NickDelta
Dec 10 '18 at 15:32
$begingroup$
What are you having trouble with? Attaching a T to each P seems to me like what you were doing and makes sure you don't have two Ps together.
$endgroup$
– Ross Millikan
Dec 10 '18 at 15:43
$begingroup$
Now i got it that phrase is equivalent to $frac{7!}(4!}$ . Very nice answer, thanks once again.
$endgroup$
– NickDelta
Dec 10 '18 at 15:47
add a comment |
$begingroup$
You're right I was missing a lot of configurations with my thought. Your results are correct but sorry for my ignorance but I didn't understood your thought. I'm not a native English speaker and understanding maths in English is even more difficult than speaking it. Could you please be a little more analytic ?
$endgroup$
– NickDelta
Dec 10 '18 at 15:32
$begingroup$
What are you having trouble with? Attaching a T to each P seems to me like what you were doing and makes sure you don't have two Ps together.
$endgroup$
– Ross Millikan
Dec 10 '18 at 15:43
$begingroup$
Now i got it that phrase is equivalent to $frac{7!}(4!}$ . Very nice answer, thanks once again.
$endgroup$
– NickDelta
Dec 10 '18 at 15:47
$begingroup$
You're right I was missing a lot of configurations with my thought. Your results are correct but sorry for my ignorance but I didn't understood your thought. I'm not a native English speaker and understanding maths in English is even more difficult than speaking it. Could you please be a little more analytic ?
$endgroup$
– NickDelta
Dec 10 '18 at 15:32
$begingroup$
You're right I was missing a lot of configurations with my thought. Your results are correct but sorry for my ignorance but I didn't understood your thought. I'm not a native English speaker and understanding maths in English is even more difficult than speaking it. Could you please be a little more analytic ?
$endgroup$
– NickDelta
Dec 10 '18 at 15:32
$begingroup$
What are you having trouble with? Attaching a T to each P seems to me like what you were doing and makes sure you don't have two Ps together.
$endgroup$
– Ross Millikan
Dec 10 '18 at 15:43
$begingroup$
What are you having trouble with? Attaching a T to each P seems to me like what you were doing and makes sure you don't have two Ps together.
$endgroup$
– Ross Millikan
Dec 10 '18 at 15:43
$begingroup$
Now i got it that phrase is equivalent to $frac{7!}(4!}$ . Very nice answer, thanks once again.
$endgroup$
– NickDelta
Dec 10 '18 at 15:47
$begingroup$
Now i got it that phrase is equivalent to $frac{7!}(4!}$ . Very nice answer, thanks once again.
$endgroup$
– NickDelta
Dec 10 '18 at 15:47
add a comment |
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$begingroup$
Now, I believe that I know what you mean when you say "circle square", but keep in mind that this is a math forum. We will think of "square" as a rectangle with equal sides before we think of "square" as a plaza.
$endgroup$
– Arthur
Dec 10 '18 at 15:06
$begingroup$
Didn't have that thought when I was translating, I'm going to edit it and put plaza to avoid confusion.
$endgroup$
– NickDelta
Dec 10 '18 at 15:08
$begingroup$
To be clear, in our circular plaza, do we care which direction north is? I.e. if we label the cars $1,2,3,dots$ and we start listing which car parked where from the northernmost space going clockwise is the parking sequence $(1,2,3,dots,9,10)$ considered the same or different than the parking sequence $(2,3,4,dots,10,1)$?
$endgroup$
– JMoravitz
Dec 10 '18 at 15:14
$begingroup$
JMoravitz I think it's considered the same since we're talking for circle permutations and not simple ones
$endgroup$
– NickDelta
Dec 10 '18 at 15:17
1
$begingroup$
As for part (b), notice that the if it is not true that no permanent residents are adjacent then it must be true that at least two permanent residents are adjacent.
$endgroup$
– JMoravitz
Dec 10 '18 at 15:21