Necklace combinatorics problem











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If $n$ is the number of necklaces which can be formed using $17$ identical pearls and two identical diamonds and similarly $m$ is the number of necklaces which can be formed using $17$ identical pearls and $2$ different diamonds, then the value of $m$ and $n$ is?






combinatorics permutations







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  • What have you done so far? Where are you stuck? Can you figure out the problem if $17$ was a smaller number like $3$ or $4$, for example? Try to extend that logic.
    – астон вілла олоф мэллбэрг
    Nov 21 at 4:09












  • Well I am not sure how to start
    – Saumil Sood
    Nov 21 at 4:10










  • In both questions, it is about where to place the diamonds. For example, let us call the diamonds $D$ (for the first question, where they are the same). Draw the skeleton of a necklace, and find out in how many ways can you place the two diamonds. The pearls will have to go in the other places, right?
    – астон вілла олоф мэллбэрг
    Nov 21 at 4:13






  • 1




    Very good. In how many ways can you select two places? Remember, we are talking about places in a necklace, which loops back into itself, so only the distance between the two diamonds matters.
    – астон вілла олоф мэллбэрг
    Nov 21 at 4:19















up vote
0
down vote

favorite












If $n$ is the number of necklaces which can be formed using $17$ identical pearls and two identical diamonds and similarly $m$ is the number of necklaces which can be formed using $17$ identical pearls and $2$ different diamonds, then the value of $m$ and $n$ is?






combinatorics permutations







share|cite|improve this question
























  • What have you done so far? Where are you stuck? Can you figure out the problem if $17$ was a smaller number like $3$ or $4$, for example? Try to extend that logic.
    – астон вілла олоф мэллбэрг
    Nov 21 at 4:09












  • Well I am not sure how to start
    – Saumil Sood
    Nov 21 at 4:10










  • In both questions, it is about where to place the diamonds. For example, let us call the diamonds $D$ (for the first question, where they are the same). Draw the skeleton of a necklace, and find out in how many ways can you place the two diamonds. The pearls will have to go in the other places, right?
    – астон вілла олоф мэллбэрг
    Nov 21 at 4:13






  • 1




    Very good. In how many ways can you select two places? Remember, we are talking about places in a necklace, which loops back into itself, so only the distance between the two diamonds matters.
    – астон вілла олоф мэллбэрг
    Nov 21 at 4:19













up vote
0
down vote

favorite









up vote
0
down vote

favorite











If $n$ is the number of necklaces which can be formed using $17$ identical pearls and two identical diamonds and similarly $m$ is the number of necklaces which can be formed using $17$ identical pearls and $2$ different diamonds, then the value of $m$ and $n$ is?






combinatorics permutations







share|cite|improve this question















If $n$ is the number of necklaces which can be formed using $17$ identical pearls and two identical diamonds and similarly $m$ is the number of necklaces which can be formed using $17$ identical pearls and $2$ different diamonds, then the value of $m$ and $n$ is?






combinatorics permutations




combinatorics permutations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 at 5:06









David G. Stork

9,48721232




9,48721232










asked Nov 21 at 4:07









Saumil Sood

93




93












  • What have you done so far? Where are you stuck? Can you figure out the problem if $17$ was a smaller number like $3$ or $4$, for example? Try to extend that logic.
    – астон вілла олоф мэллбэрг
    Nov 21 at 4:09












  • Well I am not sure how to start
    – Saumil Sood
    Nov 21 at 4:10










  • In both questions, it is about where to place the diamonds. For example, let us call the diamonds $D$ (for the first question, where they are the same). Draw the skeleton of a necklace, and find out in how many ways can you place the two diamonds. The pearls will have to go in the other places, right?
    – астон вілла олоф мэллбэрг
    Nov 21 at 4:13






  • 1




    Very good. In how many ways can you select two places? Remember, we are talking about places in a necklace, which loops back into itself, so only the distance between the two diamonds matters.
    – астон вілла олоф мэллбэрг
    Nov 21 at 4:19


















  • What have you done so far? Where are you stuck? Can you figure out the problem if $17$ was a smaller number like $3$ or $4$, for example? Try to extend that logic.
    – астон вілла олоф мэллбэрг
    Nov 21 at 4:09












  • Well I am not sure how to start
    – Saumil Sood
    Nov 21 at 4:10










  • In both questions, it is about where to place the diamonds. For example, let us call the diamonds $D$ (for the first question, where they are the same). Draw the skeleton of a necklace, and find out in how many ways can you place the two diamonds. The pearls will have to go in the other places, right?
    – астон вілла олоф мэллбэрг
    Nov 21 at 4:13






  • 1




    Very good. In how many ways can you select two places? Remember, we are talking about places in a necklace, which loops back into itself, so only the distance between the two diamonds matters.
    – астон вілла олоф мэллбэрг
    Nov 21 at 4:19
















What have you done so far? Where are you stuck? Can you figure out the problem if $17$ was a smaller number like $3$ or $4$, for example? Try to extend that logic.
– астон вілла олоф мэллбэрг
Nov 21 at 4:09






What have you done so far? Where are you stuck? Can you figure out the problem if $17$ was a smaller number like $3$ or $4$, for example? Try to extend that logic.
– астон вілла олоф мэллбэрг
Nov 21 at 4:09














Well I am not sure how to start
– Saumil Sood
Nov 21 at 4:10




Well I am not sure how to start
– Saumil Sood
Nov 21 at 4:10












In both questions, it is about where to place the diamonds. For example, let us call the diamonds $D$ (for the first question, where they are the same). Draw the skeleton of a necklace, and find out in how many ways can you place the two diamonds. The pearls will have to go in the other places, right?
– астон вілла олоф мэллбэрг
Nov 21 at 4:13




In both questions, it is about where to place the diamonds. For example, let us call the diamonds $D$ (for the first question, where they are the same). Draw the skeleton of a necklace, and find out in how many ways can you place the two diamonds. The pearls will have to go in the other places, right?
– астон вілла олоф мэллбэрг
Nov 21 at 4:13




1




1




Very good. In how many ways can you select two places? Remember, we are talking about places in a necklace, which loops back into itself, so only the distance between the two diamonds matters.
– астон вілла олоф мэллбэрг
Nov 21 at 4:19




Very good. In how many ways can you select two places? Remember, we are talking about places in a necklace, which loops back into itself, so only the distance between the two diamonds matters.
– астон вілла олоф мэллбэрг
Nov 21 at 4:19










1 Answer
1






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0
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Use Burnside's Lemma.




If $G$ is a finite group of permutations on a set $S$, then the number of orbits of $G$ on $S$ is
begin{align}
frac{1}{|G|}sum_{g in G}|operatorname{fix}(g)|.
end{align}




Observe there are $binom{19}{2}=171$ possible designs (of course, some of them are equivalent by rotation and reflection). Let $G = D_{38}$ act on the $171$ necklace designs. Observe we see that
begin{matrix}
text{element} & text{Number of Designs Fixed by Element}\
e & 171\
r^k, k=1,ldots, 18 & 0\
f & 9\
fr^k, k=1,ldots, 18 & 9
end{matrix}
Hence by Burnside's lemma, we see that
begin{align}
frac{1}{38}(171+9+9*18) = 9.
end{align}



Indeed, one possible configuration is that the two diamonds are $1$ apart on one side and $16$ apart on the other side or $3$ apart on one side and $14$ apart on the other side...until $17$ apart on one side and $0$ apart on the other side. Hence $9$ configurations.






share|cite|improve this answer





















  • Isn't this overkill...
    – YiFan
    Nov 21 at 4:56










  • @YiFan Yes. But my last paragraph gives a simple proof.
    – Jacky Chong
    Nov 21 at 4:57










  • I don't know this Theorem, can you tell any other method. Thanks
    – Saumil Sood
    Nov 21 at 14:16











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up vote
0
down vote













Use Burnside's Lemma.




If $G$ is a finite group of permutations on a set $S$, then the number of orbits of $G$ on $S$ is
begin{align}
frac{1}{|G|}sum_{g in G}|operatorname{fix}(g)|.
end{align}




Observe there are $binom{19}{2}=171$ possible designs (of course, some of them are equivalent by rotation and reflection). Let $G = D_{38}$ act on the $171$ necklace designs. Observe we see that
begin{matrix}
text{element} & text{Number of Designs Fixed by Element}\
e & 171\
r^k, k=1,ldots, 18 & 0\
f & 9\
fr^k, k=1,ldots, 18 & 9
end{matrix}
Hence by Burnside's lemma, we see that
begin{align}
frac{1}{38}(171+9+9*18) = 9.
end{align}



Indeed, one possible configuration is that the two diamonds are $1$ apart on one side and $16$ apart on the other side or $3$ apart on one side and $14$ apart on the other side...until $17$ apart on one side and $0$ apart on the other side. Hence $9$ configurations.






share|cite|improve this answer





















  • Isn't this overkill...
    – YiFan
    Nov 21 at 4:56










  • @YiFan Yes. But my last paragraph gives a simple proof.
    – Jacky Chong
    Nov 21 at 4:57










  • I don't know this Theorem, can you tell any other method. Thanks
    – Saumil Sood
    Nov 21 at 14:16















up vote
0
down vote













Use Burnside's Lemma.




If $G$ is a finite group of permutations on a set $S$, then the number of orbits of $G$ on $S$ is
begin{align}
frac{1}{|G|}sum_{g in G}|operatorname{fix}(g)|.
end{align}




Observe there are $binom{19}{2}=171$ possible designs (of course, some of them are equivalent by rotation and reflection). Let $G = D_{38}$ act on the $171$ necklace designs. Observe we see that
begin{matrix}
text{element} & text{Number of Designs Fixed by Element}\
e & 171\
r^k, k=1,ldots, 18 & 0\
f & 9\
fr^k, k=1,ldots, 18 & 9
end{matrix}
Hence by Burnside's lemma, we see that
begin{align}
frac{1}{38}(171+9+9*18) = 9.
end{align}



Indeed, one possible configuration is that the two diamonds are $1$ apart on one side and $16$ apart on the other side or $3$ apart on one side and $14$ apart on the other side...until $17$ apart on one side and $0$ apart on the other side. Hence $9$ configurations.






share|cite|improve this answer





















  • Isn't this overkill...
    – YiFan
    Nov 21 at 4:56










  • @YiFan Yes. But my last paragraph gives a simple proof.
    – Jacky Chong
    Nov 21 at 4:57










  • I don't know this Theorem, can you tell any other method. Thanks
    – Saumil Sood
    Nov 21 at 14:16













up vote
0
down vote










up vote
0
down vote









Use Burnside's Lemma.




If $G$ is a finite group of permutations on a set $S$, then the number of orbits of $G$ on $S$ is
begin{align}
frac{1}{|G|}sum_{g in G}|operatorname{fix}(g)|.
end{align}




Observe there are $binom{19}{2}=171$ possible designs (of course, some of them are equivalent by rotation and reflection). Let $G = D_{38}$ act on the $171$ necklace designs. Observe we see that
begin{matrix}
text{element} & text{Number of Designs Fixed by Element}\
e & 171\
r^k, k=1,ldots, 18 & 0\
f & 9\
fr^k, k=1,ldots, 18 & 9
end{matrix}
Hence by Burnside's lemma, we see that
begin{align}
frac{1}{38}(171+9+9*18) = 9.
end{align}



Indeed, one possible configuration is that the two diamonds are $1$ apart on one side and $16$ apart on the other side or $3$ apart on one side and $14$ apart on the other side...until $17$ apart on one side and $0$ apart on the other side. Hence $9$ configurations.






share|cite|improve this answer












Use Burnside's Lemma.




If $G$ is a finite group of permutations on a set $S$, then the number of orbits of $G$ on $S$ is
begin{align}
frac{1}{|G|}sum_{g in G}|operatorname{fix}(g)|.
end{align}




Observe there are $binom{19}{2}=171$ possible designs (of course, some of them are equivalent by rotation and reflection). Let $G = D_{38}$ act on the $171$ necklace designs. Observe we see that
begin{matrix}
text{element} & text{Number of Designs Fixed by Element}\
e & 171\
r^k, k=1,ldots, 18 & 0\
f & 9\
fr^k, k=1,ldots, 18 & 9
end{matrix}
Hence by Burnside's lemma, we see that
begin{align}
frac{1}{38}(171+9+9*18) = 9.
end{align}



Indeed, one possible configuration is that the two diamonds are $1$ apart on one side and $16$ apart on the other side or $3$ apart on one side and $14$ apart on the other side...until $17$ apart on one side and $0$ apart on the other side. Hence $9$ configurations.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 21 at 4:52









Jacky Chong

17.5k21128




17.5k21128












  • Isn't this overkill...
    – YiFan
    Nov 21 at 4:56










  • @YiFan Yes. But my last paragraph gives a simple proof.
    – Jacky Chong
    Nov 21 at 4:57










  • I don't know this Theorem, can you tell any other method. Thanks
    – Saumil Sood
    Nov 21 at 14:16


















  • Isn't this overkill...
    – YiFan
    Nov 21 at 4:56










  • @YiFan Yes. But my last paragraph gives a simple proof.
    – Jacky Chong
    Nov 21 at 4:57










  • I don't know this Theorem, can you tell any other method. Thanks
    – Saumil Sood
    Nov 21 at 14:16
















Isn't this overkill...
– YiFan
Nov 21 at 4:56




Isn't this overkill...
– YiFan
Nov 21 at 4:56












@YiFan Yes. But my last paragraph gives a simple proof.
– Jacky Chong
Nov 21 at 4:57




@YiFan Yes. But my last paragraph gives a simple proof.
– Jacky Chong
Nov 21 at 4:57












I don't know this Theorem, can you tell any other method. Thanks
– Saumil Sood
Nov 21 at 14:16




I don't know this Theorem, can you tell any other method. Thanks
– Saumil Sood
Nov 21 at 14:16


















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