Latest TCS Aptitude Question SOLUTION: In how many rearrangements of word AMAZED is the letter 'E' positioned in between the 2 'A's (not necessarily flanked)? 1)24 2)72 3)120 4)240

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(#M40012690) TCS Numerical Ability question Keep an EYE

In how many rearrangements of word AMAZED is the letter 'E' positioned in between the 2 'A's
(not necessarily flanked)?
1)24
2)72
3)120
4)240

Asked In TCS Aparna Rakate (11 years ago)

Unsolved Read Solution (16)

Is this Puzzle helpful?   (16)   (9) Submit Your Solution Permutation and Combination

Read Solution (16) :

snehil  (11 years ago)

sorry its not necessary to be flanked
(AEA)MZD=4!=24
(AEMA)ZD=3!*2!=12
(AEMDA)Z=2!*3!=12
(AEMDZA)=4!=24
72 ANS

Is this solution Helpfull? Yes (24) | No (8)  

Vikrant Varshney  (11 years ago)

ans:3)120
sol: at the 1st and last(sixth) place E can't be placed
for 2nd and 5th place E(based on condition)....2*4!=48
for 3rd and 4th place E........................2*(3*3*2*2)=72 or
2*[(4!-3!)*2!]=72 (in other way can do this but both wil lgive same ans 72)
so total arrangements=48+72=120

Is this solution Helpfull? Yes (13) | No (9)  

aa  (11 years ago)

In any rearrangement of the word, consider only the positions of the letters A, A and E.

These can be as A A E, A E A or E A A.

So, effectively one-third of all words will have 'E' in between the two 'A's.

The total number of rearrangements are = 360.

One-third of 360 is 120.

Is this solution Helpfull? Yes (9) | No (3)  

Vikrant Varshney  (11 years ago)

bcz flank is not necessary so only AEA will not consider. there may be AMEAZD.
there is also E is between 2A's.
so ans will be 120

Is this solution Helpfull? Yes (7) | No (7)  

rocks  (11 years ago)

(aea)zmd 4!=24

Is this solution Helpfull? Yes (5) | No (8)  

taruna  (11 years ago)

we would place aea in a group and rest will be arranged as independent entities
so the ans is 4!

Is this solution Helpfull? Yes (4) | No (5)  

NEETU MORWANI  (11 years ago)

if anyone can elaborate more on this???

Is this solution Helpfull? Yes (3) | No   

Rajesh Kumar Sethi  (11 years ago)

According to the question
E positioned between 2 A

take AEA as 1 part.
then total 4 parts. then there will be 4 places.
place 1 can be arranged in 4 ways
place 2 can be arranged in 3 ways
place 3 can be arranged in 2 ways
place 4 can be arranged in 1 ways
so total 4! i.e. 4*3*2*1 = 24

Is this solution Helpfull? Yes (3) | No (1)  

abhijeet patil  (11 years ago)

120

(aea)zmd=48
azmeda=48
zameda=12
amedaz=12
total=120

Is this solution Helpfull? Yes (2) | No (3)  

pradeep  (11 years ago)

ans is 4*6=24

Is this solution Helpfull? Yes (2) | No (4)  

noor alam  (11 years ago)

in this question E should be in between two A like 'AEA' considered it as a single element then total no.of alphabates is 4
it can be arranged 4!=24
nd AEA can be arranged 3!=6 in which A come two time so 6/2!=3
now total arrangement is 24*3=72 ans.

Is this solution Helpfull? Yes (2) | No (1)  

venkatesh bollam  (11 years ago)

please elaborate it na unable to catch it

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Ritesh Tiwari  (11 years ago)

friend's it should be 4!

Is this solution Helpfull? Yes | No (2)  

preeti arya  (11 years ago)

(AEA)taken as 1 so total letters in (AEA)ZMD are 4
combinations possible are 4!=24.

Is this solution Helpfull? Yes | No (2)  

snehil  (11 years ago)

take AEA as a unit so total way = 4!=24

Is this solution Helpfull? Yes | No (1)  

srikhakollu bharghav  (11 years ago)

keep aea as one set and the terms aea,m,z,d can be arranged in 4! ways...so answer is 24...

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