How many different possible arrangements can be obtained from the lett

Question

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

Lucky2783 · Accepted Answer

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

A-B = 900

Awli · Answer

Could you explain me in detail why B is 6! / 2! ?

Thank you.

Lucky2783 · Answer

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 !

OptimusPrepJanielle · Answer

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). -- Optimus Prep's GMAT On Demand course for only $299 covers all verbal and quant. concepts in detail. Visit the following link to get your 7 days free trial account: https://www.optimus-prep.com/gmat-on-demand-course

Awli · Answer

Yes, it's been very helpful. Thank you very much!

Vinayprajapati · Answer

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

Kinshook · Answer

Asked: 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?

Number of arrangements = All possible arrangements - Arrangements with II together 
= 7!/2!2! - 6!/2! = 900

IMO C

Posted from my mobile device

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How many different possible arrangements can be obtained from the lett

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