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A key ring has 7 keys. How many different ways can they be a [#permalink]
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30 Sep 2012, 17:27
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A key ring has 7 keys. How many different ways can they be arranged? A. 6 B. 7 C. 5! D. 6! E. 7! I would like to argue against OA. 7 => 6! circular permutations. Therefore, for a keyring  arrangements = 6!/2 (clockwise/anticlockwise doesn't matter)
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Re: A key ring has 7 keys. How many different ways can they be a [#permalink]
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01 Oct 2012, 03:18
voodoochild wrote: A key ring has 7 keys. How many different ways can they be arranged?
a 6 b 7 c 5! d 6! e 7!
I would like to argue against OA. 7 => 6! circular permutations. Therefore, for a keyring  arrangements = 6!/2 (clockwise/anticlockwise doesn't matter) Hi voodoochild, Why wouldn't the direction matter? It's a circular arrangement and you will look at the keyring from one side. Say there are three rings, A, B and C.. .ABC and CBA are two different arrangements, even in a circular arrangement.
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Re: A key ring has 7 keys. How many different ways can they be a [#permalink]
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01 Oct 2012, 03:42
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voodoochild wrote: A key ring has 7 keys. How many different ways can they be arranged?
a 6 b 7 c 5! d 6! e 7!
I would like to argue against OA. 7 => 6! circular permutations. Therefore, for a keyring  arrangements = 6!/2 (clockwise/anticlockwise doesn't matter) Think of choosing one specific key and placing it on the key ring. The remaining 6 keys can now be arranged in 6! ways. Once you have placed one key, you created a row (straighten the circle), so the regular rule for permutations apply. It doesn't matter if you slide the keys clockwise or counter clockwise, they have the same relative position each to the other. For example, just for 3 keys, ABC and ACB are different arrangements. Mirror image arrangements are different, because you don't flip your key ring. At least I think this is what is assumed. In the case of arrangements around a round table, it is obvious. For sure, you cannot flip your table and the people with it.
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Re: A key ring has 7 keys. How many different ways can they be a [#permalink]
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01 Oct 2012, 04:18
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Thanks Eva. I think that we need to apply circular permutations for necklace. Isn't it? http://www.transtutors.com/mathhomewor ... tions.aspxI read this concept while preparing for Indian Engineering entrance examination.... not sure....



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Re: A key ring has 7 keys. How many different ways can they be a [#permalink]
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01 Oct 2012, 07:58
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voodoochild wrote: Thanks Eva. I think that we need to apply circular permutations for necklace. Isn't it? http://www.transtutors.com/mathhomewor ... tions.aspxI read this concept while preparing for Indian Engineering entrance examination.... not sure.... Necklace, with spherical beads, can be flipped, so symmetrical arrangements can be considered identical. But, if not explicitly stated, I think you shouldn't assume anything like this.
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Re: A key ring has 7 keys. How many different ways can they be a [#permalink]
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01 Dec 2013, 23:32
This question was ambiguous so I got it wrong. I interpret "keys on a ring" meaning the relative ordering can't change, only moved around the ring. That means either 1 ordering (not an option here), or something where I consider the ordering lefttoright as the keys hang on the ring, but I can move the leftmost key around the loop to put it in the rightmost position, as follows:
1 2 3 4 5 6 7 2 3 4 5 6 7 1 ... 6 7 1 2 3 4 5 7 6 1 2 3 4 5



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Re: A key ring has 7 keys. How many different ways can they be a [#permalink]
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19 Jan 2014, 04:58
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if you are like me who chosed 7! [as is evident 42% of us chose this] then please refer to below link for an awesome explanation https://www.youtube.com/watch?v=tn6sc0q2ILg
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Re: A key ring has 7 keys. How many different ways can they be a [#permalink]
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08 Mar 2017, 23:08
voodoochild wrote: A key ring has 7 keys. How many different ways can they be arranged?
A. 6 B. 7 C. 5! D. 6! E. 7!
I would like to argue against OA. 7 => 6! circular permutations. Therefore, for a keyring  arrangements = 6!/2 (clockwise/anticlockwise doesn't matter) Official solution from Veritas Prep. In a permutations problem that involves a circular arrangement (such as keys on a ring, or people seated around a circular table), the number of permutations is equal to (N  1)!. This is because there is no beginning or end to the ring (or no left end or right end). Consider the threeitem arrangement of A, B, C. In a row, ABC is different from BCA, but if they're in a circle they're the same arrangement: B is between A and C, A is between C and B, etc. So in this problem, with 7 keys, the calculation is (7  1)!, which is 6! which equals 720.
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