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Two keys are to be put into already exisiting 5 keys in a

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Two keys are to be put into already exisiting 5 keys in a [#permalink] New post 26 Apr 2006, 06:19
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
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 [#permalink] New post 26 Apr 2006, 06:36
is it 1/6?

The first key can be put anywhere. after that, there a 6 spots the 2nd key could be put in.
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 [#permalink] New post 26 Apr 2006, 10:46
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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 [#permalink] New post 26 Apr 2006, 17:51
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
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 [#permalink] New post 26 Apr 2006, 18:34
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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 [#permalink] New post 26 Apr 2006, 19:04
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. :oops:
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 [#permalink] New post 27 Apr 2006, 01:04
I got 2/9

what is OA?
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 [#permalink] New post 27 Apr 2006, 06:12
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.
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 [#permalink] New post 27 Apr 2006, 07:03
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.
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 [#permalink] New post 27 Apr 2006, 08:02
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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 [#permalink] New post 27 Apr 2006, 08:07
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...
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 [#permalink] New post 27 Apr 2006, 08:12
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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 [#permalink] New post 27 Apr 2006, 08:22
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 1-5 in different positiosns. Similarly, Key A and Key B can move to different locations for a fixed positioning of keys 1-5.

That's my thought on this. Bascially, if Keys1-5 are fixed and cannot be shifted, then yes, I agree with your approach. But if it means I can change the arrangement of Keys1-5, then I'll stick with my approach.

Of course, I can be wrong, no one is always right. :wink:
It's a pretty interesting question, I must say.
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 [#permalink] New post 27 Apr 2006, 08:33
ywilfred wrote:
no one is always right. :wink:


ywilfred, I totally agree :-D

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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 [#permalink] New post 27 Apr 2006, 08:36
ccax wrote:
ywilfred wrote:
no one is always right. :wink:


ywilfred, I totally agree :-D

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 :-D
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 [#permalink] New post 27 Apr 2006, 09:20
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
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 [#permalink] New post 27 Apr 2006, 09:25
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
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 [#permalink] New post 27 Apr 2006, 12:29
A little more discussions for your reference. :)
http://www.gmatclub.com/phpbb/viewtopic.php?t=23035
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 [#permalink] New post 28 Apr 2006, 01:05
HongHu wrote:
A little more discussions for your reference. :)
http://www.gmatclub.com/phpbb/viewtopic.php?t=23035


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 key-ring and you can't take them out of the key ring and do some re-arranging.

Still, it's good discussion and I agree with the answer of 1/3.
  [#permalink] 28 Apr 2006, 01:05
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