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February 21st, 2019, 12:15 PM  #1 
Newbie Joined: Feb 2019 From: London Posts: 7 Thanks: 0  Cyclic Permutation?
A chain is to contain links of eight different types of material. There are going to be nine links. Eight have a circular shape with two links made of the same material while the rest are made of different materials. The ninth link has a figure of eight shape. The two links made of the same element must not connect together. In how many different ways could we assemble the chain?(Hints. Think about how you put an asymmetric ring on your finger or link your thumbs and forefingers.) $\displaystyle \frac{(91)!}{2}  \frac{(81)!}{2}$ I basically eliminated all the permutations where the links made of same material are next to each other from the total number of permutations. I'm not sure whether this is circular permutation, since not all the links have the same shape? Did I do this right? Also, how would I take into account the "figure of 8 shape" of one of the links? My assumption would be that the links adjacent to this one can be linked in different ways... For example, horizontally like all the other links, vertically in 2 ways where one of the "o" of the "8" will be used to connect with the adjacent links... Last edited by skipjack; February 21st, 2019 at 07:59 PM. 
February 21st, 2019, 09:39 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,452 Thanks: 1337 
are the links connected into a circular chain or are they left as a entire length of chain?

February 21st, 2019, 10:18 PM  #3 
Newbie Joined: Feb 2019 From: London Posts: 7 Thanks: 0  
February 21st, 2019, 10:51 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,452 Thanks: 1337 
Ok so to distill this problem down. We have 9 total links forming a circular chain. Two of the links are indistinguishable, the other 7 are distinct. One way or another these links are attached to two neighbors each. It is stipulated that the two links of the same material cannot be neighbors. This sound right or is there some further property of the figure 8 link I'm missing? 
February 21st, 2019, 11:08 PM  #5  
Newbie Joined: Feb 2019 From: London Posts: 7 Thanks: 0  Quote:
Since they didn't specifically mention whether the figure 8 link can be linked with adjacent links in different ways... i guess i was overcomplicating things.  
February 22nd, 2019, 12:45 AM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 2,452 Thanks: 1337 
Ignoring the rule about the 2 indistinct links not being adjacent for a moment there are a total of $n_t = \dfrac{\dbinom{9}{2}7!}{9}= 20160$ arrangements possible. There will be $n_2=\dfrac{9 \cdot 7!}{9} = 7! = 5040$ arrangements that violate the rule So there are $n_{ok} = 20160  5040 = 15120$ acceptable arrangements 
February 22nd, 2019, 10:14 AM  #7  
Newbie Joined: Feb 2019 From: London Posts: 7 Thanks: 0  Quote:
Thanks! So was I wrong on the circular/cyclic permutation part? Also could you explain how you came up with the arrangements?  
February 22nd, 2019, 01:22 PM  #8  
Senior Member Joined: Sep 2015 From: USA Posts: 2,452 Thanks: 1337  Quote:
Quote:
In the first case we note that we can first choose 2 slots of 9 for the indistinguishable links. Then we can fill the remaining 7 slots however and that's the 7! Then for the unacceptable cases we note there are 9 pairs of adjacent links. 1 and 9 are considered adjacent. We fill those with the indistinguishable links and can choose the remaining 7 links 7! ways. We then take the difference to determine the number of acceptable arrangements.  

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