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# Two keys are to be put on a circular key chain that already

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Two keys are to be put on a circular key chain that already [#permalink]

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24 Jan 2005, 22:33
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Two keys are to be put on a [c]circular[/b] key chain that already has 5 keys. What is the probability of they being next to each other?

Last edited by HongHu on 04 Feb 2005, 12:55, edited 1 time in total.
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25 Jan 2005, 10:21
Total events i.e. picking 2 spaces from 7 spaces = 7C2

Favorable events, consider 2 keys as 1 unit, then you have 6 units and 6 spaces i.e. 6C1 ways to put this one unit = 6

P(e) = 6/21 = 2/7
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25 Jan 2005, 10:23
That's not correct. Where do you get seven spaces?
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25 Jan 2005, 10:25
HongHu wrote:
That's not correct. Where do you get seven spaces?

7 keys have to have 7 places where they can be placed. Do u have the OA ? It is just like having 2 ppl sit together in a row of 7.
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25 Jan 2005, 10:26
It's a circle, not a row though. And the five keys are already in the chain.
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25 Jan 2005, 11:20
ans should be 1/2.

there are only two possibility either being puton the same side or not.
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25 Jan 2005, 15:59
If the Key chain was such that keys are in a row:

By grouping two new keys together, and checking various possible combinations.
P(next to each other) = (6!*2)/ 7!
= 2/7

If the Key chain was such that keys are in a circle:
P(next to each other) = (6-1)!*(2)/ (7-1)!
= 1/3

If the Key chain was such that other 5 keys can't be removed from their positions, the question does not make too much sense to me.
May be somebody else can explain.
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25 Jan 2005, 16:03
It's actually common sense, I think. After key 6 is added to the chain, there's only six positions key 7 can take, and two of them are next to key 6. So posibility =2/6/=1/3.
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03 Feb 2005, 07:20
Nocilis,

A basic question.Based on your 1st scenario(in a row),why do you consider:
Successful Cases:(6!*2)
++++
Should it not be (5!)2 as we have 5 keys and not 6?Thank you.Rgds,

Anna

nocilis wrote:
If the Key chain was such that keys are in a row:

By grouping two new keys together, and checking various possible combinations.
P(next to each other) = (6!*2)/ 7!
= 2/7

If the Key chain was such that keys are in a circle:
P(next to each other) = (6-1)!*(2)/ (7-1)!
= 1/3

If the Key chain was such that other 5 keys can't be removed from their positions, the question does not make too much sense to me.
May be somebody else can explain.

_________________

We can crack the exam together

Intern
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04 Feb 2005, 10:23
Please tell me if I'm missing something:

Ways two keys can be together:

K1 K2 _ _ _ _ _
K2 K1 _ _ _ _ _
_ _ _ _ _ K1 K2
_ _ _ _ _ K2 K1
K1 _ _ _ _ _ K2
K2 _ _ _ _ _ K1

So total ways K1 and K2 can be adjacent = 4
Total possible ways K1 and K2 can be on the chain = 6
so prob( K1 and K2 are adjacent) = 4/6 = 2/3

What's the gap in my logic? Thanks!
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04 Feb 2005, 10:25
Why are you keeping the other five keys together?
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04 Feb 2005, 12:41
Well key chain need not be a ring. I have a cylindrical open ended key chain.
Anirban
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04 Feb 2005, 12:55
I'm modifying the question to specify it is a circular key chain.
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04 Feb 2005, 16:06
I would do this problem this way:
There are 10 ways two keys could be together.
There are 30 ways that 2 keys could be added to the chain:
There are 5 ways how 6th key could be added to the chain and once it is added, there are 6 ways how 7th key could be added, thus 6*5=30
P=10/30=1/3
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04 Feb 2005, 16:43
Tha sum gets more clearer after specifying that the key chain is circular. Good qtn though
[#permalink] 04 Feb 2005, 16:43
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# Two keys are to be put on a circular key chain that already

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