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 Post subject: permutationPosted: Wed, 16 May 2012 20:39:49 UTC
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Joined: Wed, 16 May 2012 20:27:22 UTC
Posts: 3
I am really finding some trouble in solving a particular permutation problem:-

In how many ways can 4 rings be worn in 3 fingures ?
I have checked the anwers and it says 3^4 since each ring can be worn by 3 ways and there are 4 rings !

But my answers are still very complicated and i even manually tried to calculate the number of ways and they were way beyound 3^4

For Eg:-If we simplify the above problem and say In how many ways can 2 rings be worn in 3 fingures ?
but see below my results where the answer is comming out to be 12 and since it is a case of permutation i believe the order in which the rings are worn should be given importance too.
Let r1 and r2 be the two rings:-

{r1,r2,0} {r1,0,r2} {(r1r2),0,0} {(r2r1),0,0} {r2,r1,0} {0,(r2r1),0} {0,(r1r2),0}{0,r1,r2}{r2,0,r1}{0,r2,r1} {0,0,(r1r2)} {0,0,(r2r1)}

Any help in this matter will really be useful.

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 Post subject: Re: permutationPosted: Wed, 16 May 2012 22:57:41 UTC
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Joined: Fri, 1 Jul 2011 01:17:26 UTC
Posts: 467
shobhit wrote:
I am really finding some trouble in solving a particular permutation problem:-

In how many ways can 4 rings be worn in 3 fingures ?
I have checked the anwers and it says 3^4 since each ring can be worn by 3 ways and there are 4 rings !

But my answers are still very complicated and i even manually tried to calculate the number of ways and they were way beyound 3^4

For Eg:-If we simplify the above problem and say In how many ways can 2 rings be worn in 3 fingures ?
but see below my results where the answer is comming out to be 12 and since it is a case of permutation i believe the order in which the rings are worn should be given importance too.
Let r1 and r2 be the two rings:-

{r1,r2,0} {r1,0,r2} {(r1r2),0,0} {(r2r1),0,0} {r2,r1,0} {0,(r2r1),0} {0,(r1r2),0}{0,r1,r2}{r2,0,r1}{0,r2,r1} {0,0,(r1r2)} {0,0,(r2r1)}

Any help in this matter will really be useful.

The statement of the problem appears ambiguous. The given answer assumes the order of the rings on a given finger does matter. You are assuming it does.

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 Post subject: Re: permutationPosted: Wed, 16 May 2012 23:16:35 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 13993
Location: Austin, TX
shobhit wrote:
I am really finding some trouble in solving a particular permutation problem:-

In how many ways can 4 rings be worn in 3 fingures ?
I have checked the anwers and it says 3^4 since each ring can be worn by 3 ways and there are 4 rings !

But my answers are still very complicated and i even manually tried to calculate the number of ways and they were way beyound 3^4

For Eg:-If we simplify the above problem and say In how many ways can 2 rings be worn in 3 fingures ?
but see below my results where the answer is comming out to be 12 and since it is a case of permutation i believe the order in which the rings are worn should be given importance too.
Let r1 and r2 be the two rings:-

{r1,r2,0} {r1,0,r2} {(r1r2),0,0} {(r2r1),0,0} {r2,r1,0} {0,(r2r1),0} {0,(r1r2),0}{0,r1,r2}{r2,0,r1}{0,r2,r1} {0,0,(r1r2)} {0,0,(r2r1)}

Any help in this matter will really be useful.

Can you tell the difference between the rings? If not, then there is absolutely no way there are more than .

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 Post subject: Re: permutationPosted: Thu, 17 May 2012 16:02:32 UTC
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Joined: Wed, 16 May 2012 20:27:22 UTC
Posts: 3
The order does matter in this case as the rings are not identical.
I found this question in basic problems for permutation where permutation of n different things is taken when the repetition is allowed ...
Not sure how this problem is related to the topic and how they found this answer (3^4):(
And also even if the rings are identical the answer wont be 3^4
bec in that case r1 and r2 will be same the answer will be 6 for the example i took for 2 rings !!

But it seems like something needs attention to this problem

When i was going to this site i could even find a post for a similer problem where the answer was given as 3^5

question is for 5 rings in 3 fingures

See 8th question on this post...
viewtopic.php?t=37881

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 Post subject: Re: permutationPosted: Thu, 17 May 2012 20:44:49 UTC
 Moderator

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 13993
Location: Austin, TX
shobhit wrote:
The order does matter in this case as the rings are not identical.
I found this question in basic problems for permutation where permutation of n different things is taken when the repetition is allowed ...
Not sure how this problem is related to the topic and how they found this answer (3^4):(
And also even if the rings are identical the answer wont be 3^4
bec in that case r1 and r2 will be same the answer will be 6 for the example i took for 2 rings !!

But it seems like something needs attention to this problem

When i was going to this site i could even find a post for a similer problem where the answer was given as 3^5

question is for 5 rings in 3 fingures

See 8th question on this post...
viewtopic.php?t=37881

Fine, say the rings are Blue, Red, Green, and Yellow.

If I have three fingers and I put the Green one on one and then the Yellow one, is that different from Yellow and then Green?

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 Post subject: Re: permutationPosted: Fri, 18 May 2012 19:10:45 UTC
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Joined: Wed, 16 May 2012 20:27:22 UTC
Posts: 3
As per my understanding it should be different.

and if the answer to this problem is 3^4 ....

Will it be possible for you to write down the the 9 cases(the possible combinations ) which will be the answer for a simpler problem with two rings (Lets say blue and green) that can be worn in 3 fingures.

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 Post subject: Re: permutationPosted: Fri, 18 May 2012 21:59:31 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 13993
Location: Austin, TX
shobhit wrote:
As per my understanding it should be different.

and if the answer to this problem is 3^4 ....

Will it be possible for you to write down the the 9 cases(the possible combinations ) which will be the answer for a simpler problem with two rings (Lets say blue and green) that can be worn in 3 fingures.

If this were my homework I might, however I'm not going to do that. If you want to work out what you think would be the solution for that case I can look it over for you, however.

As it stands the problem cannot be correct as indeed, with distinguishable rings, this is not true (there are 12 possibilities with two rings for example). I would make certain the rings actually are meant to be distinguishable.

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