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Five letters A, A, B, C, D has to be arranged such that A, A 25 May 2005, 21:57
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Five letters A, A, B, C, D has to be arranged such that A, A will not be next to each other?



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Five letters A, A, B, C, D has to be arranged such that A, A 25 May 2005, 23:02
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Vithal
isn't it 5!/2! - 4! = 60 - 24 = 36
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nope vithal. it is very difficult to explain but the difference is subtle. for example:
(i) according to the question, abcda is one arrangement but aabcd is not.
(ii) according to your approach, abcda can not be an arrangement.

will explain if any..................
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Five letters A, A, B, C, D has to be arranged such that A, A 25 May 2005, 23:22
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Vithal
isn't it
5!/2! - 4! = 60 - 24 = 36
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Your answer is perfectly right if permutations between A's don't matter. Could be true. Question doesn't specify.

But usually when I see the word "arrangement" I think ordering.
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Five letters A, A, B, C, D has to be arranged such that A, A 26 May 2005, 13:08
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Vithal
isn't it
5!/2! - 4! = 60 - 24 = 36

Your answer is perfectly right if permutations between A's don't matter. Could be true. Question doesn't specify.

But usually when I see the word "arrangement" I think ordering.
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I did not get this?? Could you please explain again? In what case it would be 36?
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Five letters A, A, B, C, D has to be arranged such that A, A 26 May 2005, 14:33
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Say you have two A's A1 and A2, Vithal's answer doesn't count A1 A2 and A2 A1 as two different permutations.
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Five letters A, A, B, C, D has to be arranged such that A, A 26 May 2005, 15:10
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5!/2!-4!...

5 spaces => A A B C D

A A is 1 unity. so you can arrange them with A A always together in 4! ways. you dont multiply by 2 because A and A are the same. the total arrangements of A A B C D are 5!/2!. here you divide by 2! because they are the same. so its total ways - ways where they are always together=ways where they are never together.
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Five letters A, A, B, C, D has to be arranged such that A, A 02 Jun 2005, 22:33
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Let's look at a simple case. How many ways can A, A, B be arranged?

Method one: 3!/2!=3
Method two: 3!-2!=4

Arrangement:
AAB
ABA
BAA

Arrangement of A1, A2, B:
A1 A2 B
A2 A1 B
A1 B A2
A2 B A1
B A1 A2
B A2 A1

Method 2 tries to get rid of the repeated case by taking out the ones when the two A's are together. However in reality even if the two As are not together there could still be repeated cases. Therefore method one is correct and method two is not correct.
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Five letters A, A, B, C, D has to be arranged such that A, A 03 Jun 2005, 08:49
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HongHu
Let's look at a simple case. How many ways can A, A, B be arranged?

Method one: 3!/2!=3
Method two: 3!-2!=4

Arrangement:
AAB
ABA
BAA

Arrangement of A1, A2, B:
A1 A2 B
A2 A1 B
A1 B A2
A2 B A1
B A1 A2
B A2 A1

Method 2 tries to get rid of the repeated case by taking out the ones when the two A's are together. However in reality even if the two As are not together there could still be repeated cases. Therefore method one is correct and method two is not correct.
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So method 1 means ans should be

5!/2! - 4! = 36

Do all agree ?
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Five letters A, A, B, C, D has to be arranged such that A, A 03 Jun 2005, 10:28
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jpv
I am sorry, If this type of question is posted earlier.

Five letters A, A, B, C, D has to be arranged such that A, A will not be next to each other?
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total number of arrangement = 5!
number of arrangements were AA are together 2*4!
answer = 5!-2*4! = 72



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