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

• 4!=24 ans......
• 11 years agoHelpfull: Yes(22) No(6)
• 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
• 11 years agoHelpfull: Yes(16) No(21)
• they mentioned directly....that A must be between two 'E' s
  So the word RSD(EAE) can be arranged in 4! ways =24
  
  
• 11 years agoHelpfull: Yes(9) No(2)
• 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
• 11 years agoHelpfull: Yes(5) No(7)
• @jayant the total no of combinations is 60
• 11 years agoHelpfull: Yes(4) No(4)
• 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.
• 11 years agoHelpfull: Yes(3) No(1)
• consider 'EAE' a single letter.. then the no. of arrangements are = 4!
• 11 years agoHelpfull: Yes(2) No(2)
• 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.
• 11 years agoHelpfull: Yes(1) No(7)
• read the question carefully its A is b/w 2 E's not only A is b/w 2 E's
• 11 years agoHelpfull: Yes(1) No(0)
• 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
• 11 years agoHelpfull: Yes(1) No(3)
• yes pooja you are right ans is 60
• 11 years agoHelpfull: Yes(1) No(1)
• 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
• 11 years agoHelpfull: Yes(1) No(1)
• we consider EAE as 1 then 6-3+1=5
  so num of rearrangement= 5!= 120
• 11 years agoHelpfull: Yes(1) No(5)
• correct sol. is 120-1= 119
• 11 years agoHelpfull: Yes(0) No(2)
• 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....
• 11 years agoHelpfull: Yes(0) No(0)
• 3 vows. so (6-3+1)!=4!=24 ain middle. so vows can arrange only in 1 way.
  so ans=24*1=24
• 11 years agoHelpfull: Yes(0) No(0)
• A is btwen e and e.rsd are one set eae are one set no ways 3(rsd)+1(eae)=4!=24
• 11 years agoHelpfull: Yes(0) No(0)
• 24
  surbhi u r r8........
• 11 years agoHelpfull: Yes(0) No(0)
• None of these
• 4 years agoHelpfull: Yes(0) No(0)