circular r-permutations of n
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My book, Discrete Mathematics And Its Applications by K.H.Rosen asks to find a formula for circular r-permutations of n people. That is, sitting of r of these people around a table when two sittings are considered equal if they look the same by rotation of the table.
As I searched, the answer must be, P(n,r)/2r but the book says something different:
The books answer.
So my problem is why? Why do we divide by r and not by 2 at the end?
And why didn't "design the head" step divide the answer by r?
combinatorics permutations
asked Dec 4 '18 at 16:23
AK 12AK 12
11
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add a comment |
$begingroup$
My book, Discrete Mathematics And Its Applications by K.H.Rosen asks to find a formula for circular r-permutations of n people. That is, sitting of r of these people around a table when two sittings are considered equal if they look the same by rotation of the table.
As I searched, the answer must be, P(n,r)/2r but the book says something different:
The books answer.
So my problem is why? Why do we divide by r and not by 2 at the end?
And why didn't "design the head" step divide the answer by r?
combinatorics permutations
asked Dec 4 '18 at 16:23
AK 12AK 12
11
$endgroup$
$begingroup$
Think this way: How many ways can we seat $r$ people around a circular table? This is $r!/r$. To seat $r$ out of $n$ people, we first choose $r$ people in $binom{n}{r}$ ways and seat them in $(r-1)!$ ways.
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– Muralidharan
Dec 4 '18 at 16:29
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"As I searched, the answer must be, P(n,r)/2r" -- Since there are r seats and rotations don't count, you would divide by r. Why would you divide by 2?
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– dcromley
Dec 4 '18 at 16:34
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I think the issue is that you're counting reflections as being equivalent, but the book isn't.
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– user3482749
Dec 4 '18 at 16:36
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I divide by 2 because the as the book says "sittings are considered same, if they are obtained from each other by rotating the table."
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– AK 12
Dec 4 '18 at 16:40
1
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"... rotating the table". That's where therfactor comes in. 2 is for reflection symmetry.
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– karakfa
Dec 4 '18 at 17:43
add a comment |
$begingroup$
My book, Discrete Mathematics And Its Applications by K.H.Rosen asks to find a formula for circular r-permutations of n people. That is, sitting of r of these people around a table when two sittings are considered equal if they look the same by rotation of the table.
As I searched, the answer must be, P(n,r)/2r but the book says something different:
The books answer.
So my problem is why? Why do we divide by r and not by 2 at the end?
And why didn't "design the head" step divide the answer by r?
combinatorics permutations
asked Dec 4 '18 at 16:23
AK 12AK 12
11
$endgroup$
My book, Discrete Mathematics And Its Applications by K.H.Rosen asks to find a formula for circular r-permutations of n people. That is, sitting of r of these people around a table when two sittings are considered equal if they look the same by rotation of the table.
As I searched, the answer must be, P(n,r)/2r but the book says something different:
The books answer.
So my problem is why? Why do we divide by r and not by 2 at the end?
And why didn't "design the head" step divide the answer by r?
combinatorics permutations
combinatorics permutations
asked Dec 4 '18 at 16:23
AK 12AK 12
11
asked Dec 4 '18 at 16:23
AK 12AK 12
11
asked Dec 4 '18 at 16:23
AK 12AK 12
11
asked Dec 4 '18 at 16:23
AK 12AK 12
11
asked Dec 4 '18 at 16:23
AK 12AK 12
11
11
$begingroup$
Think this way: How many ways can we seat $r$ people around a circular table? This is $r!/r$. To seat $r$ out of $n$ people, we first choose $r$ people in $binom{n}{r}$ ways and seat them in $(r-1)!$ ways.
$endgroup$
– Muralidharan
Dec 4 '18 at 16:29
$begingroup$
"As I searched, the answer must be, P(n,r)/2r" -- Since there are r seats and rotations don't count, you would divide by r. Why would you divide by 2?
$endgroup$
– dcromley
Dec 4 '18 at 16:34
$begingroup$
I think the issue is that you're counting reflections as being equivalent, but the book isn't.
$endgroup$
– user3482749
Dec 4 '18 at 16:36
$begingroup$
I divide by 2 because the as the book says "sittings are considered same, if they are obtained from each other by rotating the table."
$endgroup$
– AK 12
Dec 4 '18 at 16:40
1
$begingroup$
"... rotating the table". That's where therfactor comes in. 2 is for reflection symmetry.
$endgroup$
– karakfa
Dec 4 '18 at 17:43
add a comment |
$begingroup$
Think this way: How many ways can we seat $r$ people around a circular table? This is $r!/r$. To seat $r$ out of $n$ people, we first choose $r$ people in $binom{n}{r}$ ways and seat them in $(r-1)!$ ways.
$endgroup$
– Muralidharan
Dec 4 '18 at 16:29
$begingroup$
"As I searched, the answer must be, P(n,r)/2r" -- Since there are r seats and rotations don't count, you would divide by r. Why would you divide by 2?
$endgroup$
– dcromley
Dec 4 '18 at 16:34
$begingroup$
I think the issue is that you're counting reflections as being equivalent, but the book isn't.
$endgroup$
– user3482749
Dec 4 '18 at 16:36
$begingroup$
I divide by 2 because the as the book says "sittings are considered same, if they are obtained from each other by rotating the table."
$endgroup$
– AK 12
Dec 4 '18 at 16:40
1
$begingroup$
"... rotating the table". That's where therfactor comes in. 2 is for reflection symmetry.
$endgroup$
– karakfa
Dec 4 '18 at 17:43
$begingroup$
Think this way: How many ways can we seat $r$ people around a circular table? This is $r!/r$. To seat $r$ out of $n$ people, we first choose $r$ people in $binom{n}{r}$ ways and seat them in $(r-1)!$ ways.
$endgroup$
– Muralidharan
Dec 4 '18 at 16:29
$begingroup$
Think this way: How many ways can we seat $r$ people around a circular table? This is $r!/r$. To seat $r$ out of $n$ people, we first choose $r$ people in $binom{n}{r}$ ways and seat them in $(r-1)!$ ways.
$endgroup$
– Muralidharan
Dec 4 '18 at 16:29
$begingroup$
"As I searched, the answer must be, P(n,r)/2r" -- Since there are r seats and rotations don't count, you would divide by r. Why would you divide by 2?
$endgroup$
– dcromley
Dec 4 '18 at 16:34
$begingroup$
"As I searched, the answer must be, P(n,r)/2r" -- Since there are r seats and rotations don't count, you would divide by r. Why would you divide by 2?
$endgroup$
– dcromley
Dec 4 '18 at 16:34
$begingroup$
I think the issue is that you're counting reflections as being equivalent, but the book isn't.
$endgroup$
– user3482749
Dec 4 '18 at 16:36
$begingroup$
I think the issue is that you're counting reflections as being equivalent, but the book isn't.
$endgroup$
– user3482749
Dec 4 '18 at 16:36
$begingroup$
I divide by 2 because the as the book says "sittings are considered same, if they are obtained from each other by rotating the table."
$endgroup$
– AK 12
Dec 4 '18 at 16:40
$begingroup$
I divide by 2 because the as the book says "sittings are considered same, if they are obtained from each other by rotating the table."
$endgroup$
– AK 12
Dec 4 '18 at 16:40
1
1
$begingroup$
"... rotating the table". That's where the
r factor comes in. 2 is for reflection symmetry.$endgroup$
– karakfa
Dec 4 '18 at 17:43
$begingroup$
"... rotating the table". That's where the
r factor comes in. 2 is for reflection symmetry.$endgroup$
– karakfa
Dec 4 '18 at 17:43
add a comment |
1 Answer
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What you probably got the answer is the "bracelet problem", what is being asked is the "necklace problem". The difference is the bracelet is assumed to be equivalent under reflection, whereas necklace is not. see wikipedia article here
answered Dec 4 '18 at 19:10
karakfakarakfa
1,973811
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1 Answer
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1 Answer
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$begingroup$
What you probably got the answer is the "bracelet problem", what is being asked is the "necklace problem". The difference is the bracelet is assumed to be equivalent under reflection, whereas necklace is not. see wikipedia article here
answered Dec 4 '18 at 19:10
karakfakarakfa
1,973811
$endgroup$
add a comment |
$begingroup$
What you probably got the answer is the "bracelet problem", what is being asked is the "necklace problem". The difference is the bracelet is assumed to be equivalent under reflection, whereas necklace is not. see wikipedia article here
answered Dec 4 '18 at 19:10
karakfakarakfa
1,973811
$endgroup$
add a comment |
$begingroup$
What you probably got the answer is the "bracelet problem", what is being asked is the "necklace problem". The difference is the bracelet is assumed to be equivalent under reflection, whereas necklace is not. see wikipedia article here
answered Dec 4 '18 at 19:10
karakfakarakfa
1,973811
$endgroup$
What you probably got the answer is the "bracelet problem", what is being asked is the "necklace problem". The difference is the bracelet is assumed to be equivalent under reflection, whereas necklace is not. see wikipedia article here
answered Dec 4 '18 at 19:10
karakfakarakfa
1,973811
answered Dec 4 '18 at 19:10
karakfakarakfa
1,973811
answered Dec 4 '18 at 19:10
karakfakarakfa
1,973811
answered Dec 4 '18 at 19:10
karakfakarakfa
1,973811
1,973811
add a comment |
add a comment |
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$begingroup$
Think this way: How many ways can we seat $r$ people around a circular table? This is $r!/r$. To seat $r$ out of $n$ people, we first choose $r$ people in $binom{n}{r}$ ways and seat them in $(r-1)!$ ways.
$endgroup$
– Muralidharan
Dec 4 '18 at 16:29
$begingroup$
"As I searched, the answer must be, P(n,r)/2r" -- Since there are r seats and rotations don't count, you would divide by r. Why would you divide by 2?
$endgroup$
– dcromley
Dec 4 '18 at 16:34
$begingroup$
I think the issue is that you're counting reflections as being equivalent, but the book isn't.
$endgroup$
– user3482749
Dec 4 '18 at 16:36
$begingroup$
I divide by 2 because the as the book says "sittings are considered same, if they are obtained from each other by rotating the table."
$endgroup$
– AK 12
Dec 4 '18 at 16:40
1
$begingroup$
"... rotating the table". That's where the
rfactor comes in. 2 is for reflection symmetry.$endgroup$
– karakfa
Dec 4 '18 at 17:43