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Two keys are to be put into already exisiting 5 keys in a [#permalink]
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26 Apr 2006, 07:19
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Two keys are to be put into already exisiting 5 keys in a key chain. What is the probability that these two keys will be adjacent on the chain? Assume key chain to be circular.
Please explain



Manager
Joined: 23 Jan 2006
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is it 1/6?
The first key can be put anywhere. after that, there a 6 spots the 2nd key could be put in.



Manager
Joined: 20 Nov 2004
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kook44 wrote: The first key can be put anywhere. after that, there a 6 spots the 2nd key could be put in.
That's true, but because it's a key chain, there are 2 of the 6 spots that
are adjacent to the first.
So I'd go for 1/3.



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Joined: 07 Jul 2004
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Number of ways 7 keys can be hooked to the key chain = 7! = 5040
Number of ways keys can be booked such that the two keys are always adjacent = 6! = 720 (Considering both keys as 1 entity)
The probability keys will be adjacent = 720/5040 = 1/7



Manager
Joined: 21 Mar 2006
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Quote: Number of ways 7 keys can be hooked to the key chain = 7! = 5040
Number of ways keys can be booked such that the two keys are always adjacent = 6! = 720 (Considering both keys as 1 entity)
The probability keys will be adjacent = 720/5040 = 1/7
Wouldn't it be 2/7 because we need to consider the two ways that the values can be arranged (i.e. 2*6!)?



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Charlie45 wrote: Quote: Number of ways 7 keys can be hooked to the key chain = 7! = 5040
Number of ways keys can be booked such that the two keys are always adjacent = 6! = 720 (Considering both keys as 1 entity)
The probability keys will be adjacent = 720/5040 = 1/7 Wouldn't it be 2/7 because we need to consider the two ways that the values can be arranged (i.e. 2*6!)?
Yes, you're right!! I forgot to invert the two keys.



Senior Manager
Joined: 09 Mar 2006
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I got 2/9
what is OA?



Manager
Joined: 20 Nov 2004
Posts: 108

Please have a look at the image.
We're talking about a key chain that is circular.
S1  S5 are the keys that are already on the chain.
A is the first key we put into the chain. The
position doesn't matter.
After that, there are 6 slots where the second key can
be put into. From these, 2 are adjacent to A.
So the probability should be 2/6 or 1/3.
Attachments
key.jpg [ 16.22 KiB  Viewed 961 times ]



GMAT Club Legend
Joined: 07 Jul 2004
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ccax wrote: Please have a look at the image.
We're talking about a key chain that is circular.
S1  S5 are the keys that are already on the chain.
A is the first key we put into the chain. The position doesn't matter.
After that, there are 6 slots where the second key can be put into. From these, 2 are adjacent to A.
So the probability should be 2/6 or 1/3.
I still think the probability is 2/7. Your method has not taken into consideration that the 5 keys already on the key chain can have different arrangements.... let me know what you think.



Manager
Joined: 20 Nov 2004
Posts: 108

ywilfred wrote: I still think the probability is 2/7. Your method has not taken into consideration that the 5 keys already on the key chain can have different arrangements.... let me know what you think.
The picture will ALWAYS looks like this after the first key of us has
been put into the chain? The numbers next to the other 5 keys are only
a help to count a bit quicker. And besides, what's the difference it the
other 5 keys had different arrangements?



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Posts: 5043
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ccax wrote: ywilfred wrote: I still think the probability is 2/7. Your method has not taken into consideration that the 5 keys already on the key chain can have different arrangements.... let me know what you think. The picture will ALWAYS looks like this after the first key of us has been put into the chain? The numbers next to the other 5 keys are only a help to count a bit quicker. And besides, what's the difference it the other 5 keys had different arrangements?
I am assuming the keys are not identical and similar. Let's say Key 1, Key2, Key3, Key4 and Key5 are different.
Then Key1 at S1, Key2 at S2, Key3 at S3, Key4 at S4 and Key5 at S5 is different from Key1 at S2, Key2 at S3, Key3 at S1, Key4 at S5 and Key5 at S4... and so on...



Manager
Joined: 20 Nov 2004
Posts: 108

ywilfred wrote: I am assuming the keys are not identical and similar. Let's say Key 1, Key2, Key3, Key4 and Key5 are different.
But these keys aren't important for the solution of the question. What
matters is that there are only 6 slots after our first key has been
inserted, and of these, 2 are adjacent to the first key. So how could the
probability be 2/7?



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ccax wrote: ywilfred wrote: I am assuming the keys are not identical and similar. Let's say Key 1, Key2, Key3, Key4 and Key5 are different. But these keys aren't important for the solution of the question. What matters is that there are only 6 slots after our first key has been inserted, and of these, 2 are adjacent to the first key. So how could the probability be 2/7?
Why shouldn't these 5 keys be important to the solution.
Again, I'll label the 5 keys as: Key1, Key2, Key3, Key4, Key5. I'll label the two keys to be inserted as Key A, Key B.
The arrangement: Key1, KeyA, KeyB, Key2, Key3, Key4, Key5 is different from
The arrangement: Key1, KeyA, KeyB, Key3, Key2, Key4, Key 5 is different from
The arrangement: Key1, KeyB, KeyA, Key3, Key2, Key4, Key5 is different from
The arrangement: Key1, Key2, KeyA, KeyB, Key3, Key4, Key5.... and so on...
A different arrangement can be obtained by keeping Key A and Key B are a fixed location, and have keys 15 in different positiosns. Similarly, Key A and Key B can move to different locations for a fixed positioning of keys 15.
That's my thought on this. Bascially, if Keys15 are fixed and cannot be shifted, then yes, I agree with your approach. But if it means I can change the arrangement of Keys15, then I'll stick with my approach.
Of course, I can be wrong, no one is always right.
It's a pretty interesting question, I must say.



Manager
Joined: 20 Nov 2004
Posts: 108

ywilfred wrote: no one is always right.
ywilfred, I totally agree
Of course we COULD shift the other keys to different positions, but
this is a probability question. In these, we always assume that each
case is of equal probability if not otherwise stated. If not, almost
all of these probability questions wouldn't be solvable with the
information given.
If it weren't a CIRCULAR chain, then there would be 7 spots, but it IS
a circular chain, and so, there are just 6. And you won't convince me
that 2/7 is the solution unless you showed that there are 7!



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Posts: 5043
Location: Singapore

ccax wrote: ywilfred wrote: no one is always right. ywilfred, I totally agree Of course we COULD shift the other keys to different positions, but this is a probability question. In these, we always assume that each case is of equal probability if not otherwise stated. If not, almost all of these probability questions wouldn't be solvable with the information given. If it weren't a CIRCULAR chain, then there would be 7 spots, but it IS a circular chain, and so, there are just 6. And you won't convince me that 2/7 is the solution unless you showed that there are 7!
Let's wait for the OA



Manager
Joined: 27 Mar 2006
Posts: 136

ccax wrote: kook44 wrote: The first key can be put anywhere. after that, there a 6 spots the 2nd key could be put in. That's true, but because it's a key chain, there are 2 of the 6 spots that are adjacent to the first. So I'd go for 1/3.
Yeah I am getting 1/3 as well



Manager
Joined: 27 Mar 2006
Posts: 136

ywilfred wrote: ccax wrote: Please have a look at the image.
We're talking about a key chain that is circular.
S1  S5 are the keys that are already on the chain.
A is the first key we put into the chain. The position doesn't matter.
After that, there are 6 slots where the second key can be put into. From these, 2 are adjacent to A.
So the probability should be 2/6 or 1/3. I still think the probability is 2/7. Your method has not taken into consideration that the 5 keys already on the key chain can have different arrangements.... let me know what you think.
I think 1/3 is with the fact that the 5 keys have different arragements



SVP
Joined: 03 Jan 2005
Posts: 2233

A little more discussions for your reference.
http://www.gmatclub.com/phpbb/viewtopic.php?t=23035
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GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

HongHu wrote:
I think ccax's diagram explains it pretty well. But I think that's true only if there're already a bunch of keys on the keyring and you can't take them out of the key ring and do some rearranging.
Still, it's good discussion and I agree with the answer of 1/3.










