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5 letters, including two " I ", are arranged in the way that

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Senior Manager
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New post 17 May 2005, 08:19
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5 letters, including two " I ", are arranged in the way that two " I " is not next to each other. How many ways are possible?
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New post 17 May 2005, 09:30
36...

total ways 5!/2! because two letters are the same

possible ways that two I are together 4!

5!/2! - 4!=36
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New post 17 May 2005, 09:32
pb_india wrote:
4!*2!/2! = ?? how


you have 5 objects. regard the two I`s as one object. so you have 4 objetcs to order. you can do this in 4! ways.
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New post 17 May 2005, 12:21
total ways 5!/2! because two letters are the same

=> The letters are same but the objects are different. Right.
We should do 5! (total ways of arranging all 5 letters (or objects) - 4! = 72
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New post 17 May 2005, 12:46
pb_india wrote:
total ways 5!/2! because two letters are the same

=> The letters are same but the objects are different. Right.
We should do 5! (total ways of arranging all 5 letters (or objects) - 4! = 72


It shld be 5!/2!
Let's say the 2 I's are Ia, Ib
Does IaIb contribute to a diffrent outcome from IbIa?No.
5! does this kind of double counting.
Which is why you need to divide 5! by 2!.
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New post 17 May 2005, 20:25
Its is 4!*2!/2! because there are 2 I's that are changing positions with each other. Because here the order does not matter, you will have to divide by 2! in both the total cases (5!/2!) as well as in special case, where 2 I's are together (4!*2!/2!)
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New post 17 May 2005, 20:41
christoph wrote:
36...

total ways 5!/2! because two letters are the same

possible ways that two I are together 4!

5!/2! - 4!=36


This is correct It is 36
  [#permalink] 17 May 2005, 20:41
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5 letters, including two " I ", are arranged in the way that

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