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• ## Re: [2IIM CAT Prep] CAT Prep Qn : Permutation of Letters

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• 24 [AEA] M Z D     1       2  3  4 4X3X2X1 = 24 Cheers, Nitin ________________________________ From: Ascent Education CAT, Classes, Groups
Message 1 of 4 , Mar 11, 2011
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24

[AEA] M Z D
1       2  3  4

4X3X2X1 = 24

Cheers,
Nitin

From: "Ascent Education CAT, Classes, Groups" <cat@...>
To: ascent4cat@yahoogroups.com
Sent: Friday, March 11, 2011 6:22 PM
Subject: [2IIM CAT Prep] CAT Prep Qn : Permutation of Letters

Hi

Here is a relatively easy question on rearranging letters of a word.

In how many rearrangements of the word AMAZED, is the letter ‘E’ positioned in between the 2 ‘A’s (Not necessarily flanked)?
a.    24
b.    72
c.    120
d.    240

Best wishes

2IIM - CAT Classes @ Chennai, Mumbai, Bangalore
Next weekend batch @ Chennai starts Mar 20, 2011. Faculty includes IIM Graduates

• I think 120 is the answer..... Here is the explanation: See E can be positioned at 2nd, 3rd, 4th and 5th positions only. When at 2nd position:   1st position
Message 2 of 4 , Mar 13, 2011
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 I think 120 is the answer..... Here is the explanation: See E can be positioned at 2nd, 3rd, 4th and 5th positions only. When at 2nd position:   1st position is occupied by A only. then no. of permutations: 1*1*4! = 24 ways When at 3rd position:   1st & 2nd position are occupied by A and any one of D,M or  Z which  can be done in 3*2 i.e. 6 ways and last three positions can be done in 3! ways . then no. of permutations: 6*1*3! = 36 ways When at 4th position:  3*3*2*1*2! = 36 ways When at 5th position: 4!*1*1 = 24 ways So total no. of arrangements = 24+36+36+24 = 120 ways If you think this is incorrect or have any doubts, please reply to this. Regards, Kushal Singla singla_kushal@...

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