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# How many different possible arrangements can be obtained from the lett

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Intern
Joined: 08 Jan 2015
Posts: 25
How many different possible arrangements can be obtained from the lett  [#permalink]

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30 Mar 2015, 10:30
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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?

A. 360
B. 720
C. 900
D. 1800
E. 5040
Senior Manager
Joined: 07 Aug 2011
Posts: 492
GMAT 1: 630 Q49 V27
Re: How many different possible arrangements can be obtained from the lett  [#permalink]

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30 Mar 2015, 10:36
3
Awli wrote:
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?

A. 360
B. 720
C. 900
D. 1800
E. 5040

A= all possible permutations = $$\frac{7!}{2!2!}$$
B= permutations with II together = $$\frac{6!}{2!}$$

A-B = 900
Intern
Joined: 08 Jan 2015
Posts: 25
Re: How many different possible arrangements can be obtained from the lett  [#permalink]

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30 Mar 2015, 10:43
Could you explain me in detail why B is 6! / 2! ?

Thank you.
Senior Manager
Joined: 07 Aug 2011
Posts: 492
GMAT 1: 630 Q49 V27
Re: How many different possible arrangements can be obtained from the lett  [#permalink]

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30 Mar 2015, 11:06
2
Awli wrote:
Could you explain me in detail why B is 6! / 2! ?

Thank you.

yes sure.

we have to find those cases where we have at least 1 Character between two I I , if we find those cases where there is no character between two I I and subtract this number from total permutations we will be left with those cases where there are 1 or more characters between two I I .

assume II as one unit so we have total 6 characters G M A T T (II)
total ways to arrange these 6 letters is $$\frac{6!}{2!}$$ , as we have 2 Ts so we have to divide 6! by 2! .

hope the explanation is helpful to you !
SVP
Joined: 06 Nov 2014
Posts: 1865
Re: How many different possible arrangements can be obtained from the lett  [#permalink]

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31 Mar 2015, 01:50
1
2
Awli wrote:
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?

A. 360
B. 720
C. 900
D. 1800
E. 5040

There are 7 letters G, M, A, T, I, I, and T with I and T repeated twice.
Hence total arrangements/permutations of letters = 7!/(2! * 2!) = 1260.

Now, let us say that the two "I" are always together.
So now we have 6 letters G, M, A, T, II, T with T repeated twice.
Hence total arrangements/permutations of letters = 6!/(2!) = 360.

Required arrangements = 1260 - 360
= 900
Hence option (C).

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Intern
Joined: 08 Jan 2015
Posts: 25
Re: How many different possible arrangements can be obtained from the lett  [#permalink]

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31 Mar 2015, 01:56
Lucky2783 wrote:
Awli wrote:
Could you explain me in detail why B is 6! / 2! ?

Thank you.

yes sure.

we have to find those cases where we have at least 1 Character between two I I , if we find those cases where there is no character between two I I and subtract this number from total permutations we will be left with those cases where there are 1 or more characters between two I I .

assume II as one unit so we have total 6 characters G M A T T (II)
total ways to arrange these 6 letters is $$\frac{6!}{2!}$$ , as we have 2 Ts so we have to divide 6! by 2! .

hope the explanation is helpful to you !

Yes, it's been very helpful. Thank you very much!
Manager
Joined: 24 May 2013
Posts: 73
Re: How many different possible arrangements can be obtained from the lett  [#permalink]

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19 Mar 2016, 11:04
1
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?

_ _ _ _ _ _ _
Total ways for arranging without restriction = 7!/ (2!2!) { 2! is becoz of two T's and other 2! for two I's)
Restriction : atleast one character between I's = Possible ways - both I's together i.e.o character between I's

_ _ _ _ _ (I I)
Both I's Together = 6! (Assuming 2 I's as one unit) /2!(for 2 T's) * 2! (No of arrangements of 2 I's)/2! (for 2 I's)
=6!/2!

Therefore ans = 7!/ (2!2!) -6!/2! = 900
HENCE C.

A. 360
B. 720
C. 900
D. 1800
E. 5040
Non-Human User
Joined: 09 Sep 2013
Posts: 14468
Re: How many different possible arrangements can be obtained from the lett  [#permalink]

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03 Jul 2019, 13:13
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Re: How many different possible arrangements can be obtained from the lett   [#permalink] 03 Jul 2019, 13:13
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