How many different necklaces can be formed with $6$ white and $5$ red beads?
Since total number of beads is $11$ according to me it should be $\dfrac{11!}{6!5!}$ but correct answer is $21$. How does that come?
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ə̷̶̸͇̘̜́̍͗̂̄︣͟
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Faaiz Ali
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since the necklace is "circular" there are many repeated similar combinations in $\dfrac{11!}{6!5!}$ – Dylan Zammit Mar 14 '18 at 07:19
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1By my reckoning, the answer is $26$. – Angina Seng Mar 14 '18 at 08:54
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@LordSharktheUnknown I get $26$ for a bracelet, which is invariant under both rotation and reflection. However, by convention, necklaces are only invariant under rotation. – N. F. Taussig Mar 14 '18 at 13:02
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1@N.F.Taussig Then in that case, the answer is 42. – Angina Seng Mar 14 '18 at 13:12
2 Answers
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First consider $$\dfrac{11!}{5!6!}$$ as you have done. Now since the necklace is "circular" you must divide by the number of beads to remove the repeated arrangements, that is:$$\dfrac{11!}{5!6!11}$$
Also, the necklace can be flipped over, thus twice as many combinations had been counted so far, so you must divide by 2. That is:$$\dfrac{11!}{5!\cdot6!\cdot11 \cdot 2}=21$$
Dylan Zammit
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But some necklaces are preserved by the "flipping over" operation. – Angina Seng Mar 14 '18 at 07:24
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Are you referring to the cases where the beads are "symmetrical"...such as: rrwwwrr ? – Dylan Zammit Mar 14 '18 at 07:28
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2Exactly, one needs some (slightly) non-trivial application of Burnside's lemma. – Angina Seng Mar 14 '18 at 07:35
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3According to https://en.wikipedia.org/wiki/Necklace_(combinatorics), the convention in combinatorics is that a "necklace" cannot be flipped over. If it can be flipped, it is called a "bracelet". – awkward Mar 14 '18 at 13:03
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When we are calculating the circular arrangements then we do $(n-1)!$. This is the case when clockwise and anti clockwise are different.
And when clockwise and anti clockwise are same, then we use $(n-1)!/2$.
So according to the Que we apply 2 formula $$(11-1)!/2×5!×6! =21.$$
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