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

• the word {eae} can take as a one wod now total will 4 word an they can arrange in
  4! ways=24 ways
• 10 years agoHelpfull: Yes(60) No(6)
• there is a group {eae}which can be arranged in 4 ways..and remaining words can be arranged in 3! ways . so our solution is..4*3!=24
• 10 years agoHelpfull: Yes(29) No(10)
• keep [EAE]together so
  [eae]asr
  so total arngment
  4!/2!=12
  2! for the repeated e's
• 10 years agoHelpfull: Yes(9) No(11)
• I HAVE SEEN ALL THE ANS ,AND ALL OF THEM ARE DUMMY
  FIRST THIN THAT THEY ARE ASKING FOR REARRANGEMENT NOT FOR ARRANGEMENT
  therefore ans will be (arrangement - 1)
  
  there are many approach to solve this question ,i am solving by one
  soln: in all arrangement EAE will be eithere in
  1)EEA
  2)EAE
  3)AEE
  AND TOTAL ARRANGEMENT =6!/2!= (WAY OF ARRANGING 6 THING OUT OF WHICH 2 ARE ALIKE OF ONE KIND)
  
  SO REQUIRED ARRANGEMENT (IN WHICH A BETWEEN TWO E,OUT OF THREE ONLY ONE )= 360/3=120
  HENCE REARRAGEMENT WILL BE=120-1=119 (SINCE WE CANT TAKE ERASED ,HERE ALSO A BET 2 E'S)
• 10 years agoHelpfull: Yes(9) No(14)
• if we keep EAE together and assume it as single entity then we will have 4 items to rearrange so ans is
  4!=24
• 10 years agoHelpfull: Yes(6) No(1)
• RSD(EEA)
  here we treat the vowels as one letter.
  no. of ways to arrange the letters=4!
  3 vowels in which E occurs 2 times,so the vowels can be arranged in (3!/2!)=3 ways,which are(EEA),(EAE),(AEE).
  
  So,one-third of all words will have 'A' in between the two 'E's.
  So,total no of arrangements=(4!*(3!/2!))/3=24
• 10 years agoHelpfull: Yes(5) No(0)
• the word ERASED has 6 characters in total.assuming EAE as a single word.
  so now total characters =6-3+1=4 and these can be arranged in 4!ways and EAE can be arranged in 3! ways ..but EAE has two E's in it which can be arranged in 2! ways further. so the solution is 4!*3!/2!=72
• 10 years agoHelpfull: Yes(5) No(2)
• it will be 4!*2 + 2!3! + 3!2! = 72
• 10 years agoHelpfull: Yes(4) No(3)
• Consider EAE as single elements.
  then we have total 4 elements.
  so total rearrangemrnt will be 4!=24
  
• 10 years agoHelpfull: Yes(4) No(0)
• 4! is the answer
• 10 years agoHelpfull: Yes(3) No(0)
• 4!/2! that should be the answer
• 10 years agoHelpfull: Yes(3) No(1)
• -A---- 4!
  --A--- 2!3!
  ---A-- 3!2!
  ----A- 4!
  total no of ways 24+12+12+24=96 ways
• 10 years agoHelpfull: Yes(2) No(18)
• 24
  take eae as one and remaining 3 total 4
  4p4=24
• 10 years agoHelpfull: Yes(2) No(0)
• (EAE) can be placed in 4 places. ie. *RAD, R*AD, RA*D and RAD*. now RAD can be arranaged in 3! ways. so, the answer will be 4*3!= 24
• 10 years agoHelpfull: Yes(2) No(1)
• ERASED
  E A E _ _ _
  taking EAE as 1
  1+3=4
  no of ways of arranging=4!
  total no of ways of arranging the word is=4! * 3!/2!=72
• 9 years agoHelpfull: Yes(2) No(1)
• As it has been said between 2 E's 1 A should be there so the arrangement will be EAERSD,SO EAE one group which is fixed,i.e EAE=1 ,RSD can be arranged in 3 ways.(R is at=2nd,S at 3rd D at 4th place) .so it can be arranged in 4! ways(i.e 1 from EAE and 4 from rest,i.e 1*4=4! .so 4! factorial is 24
  ANS=24
  
• 10 years agoHelpfull: Yes(1) No(1)
• there will be rsd(eae) so it will be 4!=24
• 10 years agoHelpfull: Yes(1) No(0)
• EAERSD
  suppose EAE as a single entity then number of letters are 4 so they can be arranged in 4! ways.now keeping A in between two E's this can be arranged in 2! ways so 4!*2!=48
• 10 years agoHelpfull: Yes(1) No(1)
• Make group [EAE], thus remaining 3 letters + this group, totals 4 words, which can be arranged in 4! ways. (24)
• 10 years agoHelpfull: Yes(1) No(2)
• let eae be one words so total number of arrangements of four words can be done in 4! ways i.e, 24 ways :D
• 10 years agoHelpfull: Yes(1) No(1)
• EAERSD
  In this word if we fix the fist three alphabets then remaining three can be arrange in 6 ways = 6
  REAESD also 6 ways = 6 , similarly
  RSEAED also = 6
  RSDEAE also = 6
  so there are total 24 rearrangements are possible.
• 10 years agoHelpfull: Yes(0) No(0)
• takin EAE as 1 alphabet and arranging the left outs, we can do it in 4!/2! ways. the 2! is bcoz of the two identical E's.
• 10 years agoHelpfull: Yes(0) No(3)
• eae act as a one word and remaining are 3 letters which we can arrenge them in3!=6 ways
  and total we have 3+1=4 letters
  so 6*4=24
• 10 years agoHelpfull: Yes(0) No(0)
• LET EAE as a one letter,remaining 3 letters are R,S,D......so we have to arrange these four letters according to question.....4 letters are their ..no.of ways of arrenging these four letters is 4!=4*3*2=24 ans
• 10 years agoHelpfull: Yes(0) No(0)
• ans is: 4!.......here take 'EAE' as one and there is another three...But that is why here A is placed in between two similar letter ,otherwise it will be 2!*4*.
  
• 10 years agoHelpfull: Yes(0) No(2)
• the word (eae) is treated as a single entity so, the total arrangements of rsd and eae is 4!*(3!/2!)=72
• 10 years agoHelpfull: Yes(0) No(0)
• Ad we have to find Rearrangement thus
  
  Rearrangement = Arrangement -1 ..(.because ERASED is already arrange)
  thus 4!-1=23
  
• 9 years agoHelpfull: Yes(0) No(0)