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How many different possible arrangements can be obtained

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Manager
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How many different possible arrangements can be obtained [#permalink] New post 02 Oct 2006, 01:27
00:00
A
B
C
D
E

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(N/A)

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How many different possible arrangements can be obtained from the letters G, M, A, T, I, I, and T, such that there is at least one character between both I’s?
1. 360
2. 720
3. 900
4. 1,800
5. 5,040
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 [#permalink] New post 02 Oct 2006, 01:39
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my answer is D 1800.


find those permutations that have no letters between I and subtract it from the total permutation.

1. total permutation: 7!
2. permutations to subtract from 1, just treat two I's as one element. it is 6!. tricky here: since I am going to subtract it from 7!, i need to consider the permutations of 2 I's together. so this intermediate result is 2*6!.
3. 7! - 2*6! is the rest of the permutations. but there are 2 T's, so the order does not matter for T. then divide 7!-2*6! by 2.

(7!-2*6!)/2 = 1800
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 [#permalink] New post 02 Oct 2006, 05:35
Tennis ball,

answer is C 900

how to arrive to this??
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 [#permalink] New post 02 Oct 2006, 16:46
Sorry. My previous one was wrong. it is 900.

this is the calculation:

1. the total number of arrangements is :

7!/(2x2). because Ts and Is don't have effects on permutation.

2. find those that two Is are together. Just treat the 2 Is as 1.

6!/2, Ts don't have effects, so divide 6! by 2.

then 7!/4 - 6!/2 = 900.

Guess my first thinking is too complex.
  [#permalink] 02 Oct 2006, 16:46
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