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# How many different possible arrangements can be obtained

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Manager
Joined: 12 Sep 2006
Posts: 92
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How many different possible arrangements can be obtained [#permalink]  02 Oct 2006, 01:27
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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 Iâ€™s?
1. 360
2. 720
3. 900
4. 1,800
5. 5,040
VP
Joined: 25 Jun 2006
Posts: 1173
Followers: 2

Kudos [?]: 79 [0], given: 0

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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
Manager
Joined: 12 Sep 2006
Posts: 92
Followers: 1

Kudos [?]: 2 [0], given: 0

Tennis ball,

how to arrive to this??
VP
Joined: 25 Jun 2006
Posts: 1173
Followers: 2

Kudos [?]: 79 [0], given: 0

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.
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