Latest TCS Aptitude Question SOLUTION: 1. In how many rearrangements of the word ERASED, the letter 'A' is positioned in between the two 'E's ?

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(#M40025425) TCS Numerical Ability question Keep an EYE

1. In how many rearrangements of the word ERASED, the letter 'A' is positioned in between the two 'E's ?

Asked In TCS Vinodh (10 years ago)

Unsolved Read Solution (19)

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Read Solution (19) :

mehak jindal  (10 years ago)

4!=24 ans......

Is this solution Helpfull? Yes (22) | No (6)  

jayant narwani  (10 years ago)

case1:E _ _ _ _ E => 4!=24 (A can be any where in 3 blanks)
case 2:E _ _ _ E _ =>3*3!*2=36 (A can be in 3 blanks in b/w so can be arranged in 3 ways remaining 3 letters can be arranged by 3! and 2 ways of arranging E's)
case 3: E _ _ E _ _ =>2*3!*3=36
case 4: => E _ E _ _ _ => 4*3!=24


24+36+36+24=120

Is this solution Helpfull? Yes (16) | No (21)  

Arun  (10 years ago)

they mentioned directly....that A must be between two 'E' s
So the word RSD(EAE) can be arranged in 4! ways =24

Is this solution Helpfull? Yes (9) | No (2)  

surbhi rathaur  (10 years ago)

it is 60.
E_ _ _ _E =4!*(2!/2!)[arranging 2 similar E]=24
E_ _ _E_ =3*3!*(2!/2!)=18
E_ _E_ _ =2*3!*(2!/2!)=12
E_E_ _ _ =1*3!*(2!/2!)=6
24+18+12+6=60

Is this solution Helpfull? Yes (5) | No (7)  

puja  (10 years ago)

@jayant the total no of combinations is 60

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

HARSH  (10 years ago)

as asked rearranged so guys don't include erased word in ur ans.
so answer is 120-1=119
120 is same as explain by jayant narwani.

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

Rishi Raj   (10 years ago)

consider 'EAE' a single letter.. then the no. of arrangements are = 4!

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

KARTHICK K S  (10 years ago)

we have to set as EAE ==3/2!(for repitition of 'E' 2 times)
then EAE RSD==4!
therefore (4*3*2*1)*(3/2)=36.

Is this solution Helpfull? Yes (1) | No (7)  

jayant narwani  (10 years ago)

read the question carefully its A is b/w 2 E's not only A is b/w 2 E's

Is this solution Helpfull? Yes (1) | No   

ritesh sharma  (10 years ago)

for any rearrangement consider only the position of AAE
AAE,AEA,EAA
means one third of total no. having E in b/w two E's
hence , total arr.=6!/2!=360
ans=360/3=120

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

nimish sharma  (10 years ago)

yes pooja you are right ans is 60

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

utkarsh bisht  (10 years ago)

Two E‘s can be arranged in 1 way only(because Identical letters: [2!/2!])
then A can be arranged in between E‘s is oneway.
These (EAE) treated as one group(letter).
Remaining letters(RSD)& That group(letter) =4.
So Four Letters can be arranged in 4places can be done in 4! ways

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

Anuja Singh  (10 years ago)

we consider EAE as 1 then 6-3+1=5
so num of rearrangement= 5!= 120

Is this solution Helpfull? Yes (1) | No (5)  

jayant narwani  (10 years ago)

correct sol. is 120-1= 119

Is this solution Helpfull? Yes | No (2)  

akshay kapoor  (10 years ago)

result acc to me is 60 coz its clearly said that a should be between two e bt it is not said ki eae must be together....

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parvez sajjad  (10 years ago)

3 vows. so (6-3+1)!=4!=24 ain middle. so vows can arrange only in 1 way.
so ans=24*1=24

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Sk Aaqil  (10 years ago)

A is btwen e and e.rsd are one set eae are one set no ways 3(rsd)+1(eae)=4!=24

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piyush upadhyay  (10 years ago)

24
surbhi u r r8........

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Senthil kumar  (3 years ago)

None of these

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