Latest TCS Aptitude Question SOLUTION: In how many rearrangements of the word ERASED in the letter 'A' positioned in between the 2 'E's?

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(#M40024524) TCS question Keep an EYE

In how many rearrangements of the word ERASED in the letter 'A' positioned in between the 2 'E's?

Asked In TCS ashokkumar (11 years ago)

Unsolved Read Solution (24)

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Read Solution (24) :

surbhi  (11 years ago)

24
we make group of EAE as per given question
then now we have 4 letters so it can permute by 4! ways so ans is 24

Is this solution Helpfull? Yes (15) | No (10)  

NaveenKumar  (11 years ago)

4!
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 (14) | No (6)  

Nirmal  (11 years ago)

ans is 120

A _ _ _ _ _ = 5! = 120 arrangements

121 term is EAEDRS. which is the term we needed.

Is this solution Helpfull? Yes (10) | No (12)  

seetha  (11 years ago)

(EAE)_ _ _ 4!*3!/2!= 72

Is this solution Helpfull? Yes (10) | No (20)  

jayant narwani  (11 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 (10) | No (2)  

bhavna  (11 years ago)

ans will be 24.. EAE take as a single word than apply rearrangement...

Is this solution Helpfull? Yes (8) | No (7)  

kamal behl  (11 years ago)

ans is 48 bcz
EAE_ _ _ 3!=12 WAYS
_EAE_ _ 3!=12 WAYS
_ _ EAE_ 3!=12 WAYS
_ _ _EAE 3!=12WAYS
12+12+12+12=48ANS

Is this solution Helpfull? Yes (8) | No (13)  

upendraprakash  (11 years ago)

it will be 2*4!
RSD(EAE)

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

Gaurav Kumar  (11 years ago)

6C3*3!=120

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

Sruthi karuturi  (11 years ago)

4!*(2!*2!)=4!

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

Lipika Das  (11 years ago)

case 1: E A _ _ _ _ (A in 2nd)=> 4!=24
case 2: _ _ _ _ A E (A in 5th)=> 4!=24
case 3: E _ A _ _ _ or _ E A _ _ _ (A in 3rd)=>3(any one from R,S,D)*3!(remaining 3 alphabets in 3 pos)+ 3*3!=36
case 4: _ _ _ A _ E or _ _ _ A E _ (A in 4th)=>(same as case 3)=36

total= 24*2 + 36*2=120 ways

Is this solution Helpfull? Yes (4) | No   

Pushpendra Singh  (11 years ago)

according to ques arrangements can be made like(EAE)RSD..and a block of EAE and RSD can be rearranged.EAE has arrangements of 3!/2! =3 and whole (EAE)RSD can be arranged with above condintion in 4! ways=24.thus total ways =24*3=72 ans

Is this solution Helpfull? Yes (3) | No (5)  

Rohit Bohra  (11 years ago)

ans:74
(EAE)RSD
for EAE it can be arrange in 3!/2! ways=3
so whole can be arranged in (4*3*2*1)*3=72 ans

Is this solution Helpfull? Yes (3) | No (5)  

mehak jindal  (11 years ago)

145
DEAERS
A _ _ _ _ _=120
D A _ _ _ _=24
D E A E R S=120+24+1

Is this solution Helpfull? Yes (2) | No (7)  

curious coder  (11 years ago)

------.
A cannot be placed on the 2 extremes .
So A must be placed on the remaining 4 internal positions. that can be done in 4P1=4 ways.
now there are 2 E's.the E's must be placed on either side of A.that can be done in only 1 way.
so there are still 3 places vacant after placing EAE. these remaining 3 places can be filled using the 3 remaining letters in 3! ways.
So total arrangements possible = 4 * 1 * 3! = 24

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

Yati Sawhney  (11 years ago)

its 6!(2!*3)=120
because 6 letters can be arranged in 6!ways , but in this case we have 2E's thus total no of ways wud be (6!)/2!. Now the constaint we are left with is is "A shoul be positioned b/w 2 E's) . in all such arrangements A would be either to the left of both the E's or to the right of both the E's or in between them , now these three events are equally likely therefore total no of ways wud be 6!/(2!*3)

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

rakesh  (11 years ago)

ERASED: RSD+(EAE) =3+1 = 4!
EAE =2!/2! = 1
total ways = 1*4!=24

Is this solution Helpfull? Yes (1) | No   

Prerak Jain  (11 years ago)

Ans:- (6c3 * 1)* 3! = 120
6c3 is for selecting three position for EAE and they can placed in only 1 way.
Remaining 3 letter can be arranged in 3! ways.

Is this solution Helpfull? Yes (1) | No   

Surbhi khandelwal  (11 years ago)

A... 5!/2!= 60
DA...4!/2!= 12
DEAE..2!= 2
60+12+2= 74

Is this solution Helpfull? Yes | No (5)  

Pushpendra Singh  (11 years ago)

ans will be 4!*2
=48

Is this solution Helpfull? Yes | No (3)  

Yati Sawhney  (11 years ago)

7!/(2*2!)=210

Is this solution Helpfull? Yes | No (2)  

tejasvi gamini   (11 years ago)

eae can be arranged in 4ways...rsd each can be arranged in 3! ways.4*3*3!=72ways

Is this solution Helpfull? Yes | No (4)  

PRIYA NAMDEO  (11 years ago)

144 arrangements
assume EAE as a one letter remaning letters are 3 total 4 letters we have.
by permutation there are 4! ways of arranging them =24
and EAE can be arranged in 3! ways =6
24*6=144

Is this solution Helpfull? Yes | No (1)  

Guru Pranav  (11 years ago)

24 ways !!!

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