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

• 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
• 12 years agoHelpfull: Yes(24) No(8)
• 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
  
• 12 years agoHelpfull: Yes(13) No(9)
• 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.
  
• 12 years agoHelpfull: Yes(9) No(3)
• 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
• 12 years agoHelpfull: Yes(7) No(7)
• (aea)zmd 4!=24
  
• 12 years agoHelpfull: Yes(5) No(8)
• we would place aea in a group and rest will be arranged as independent entities
  so the ans is 4!
• 12 years agoHelpfull: Yes(4) No(5)
• if anyone can elaborate more on this???
• 12 years agoHelpfull: Yes(3) No(0)
• 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
• 12 years agoHelpfull: Yes(3) No(1)
• 120
  
  (aea)zmd=48
  azmeda=48
  zameda=12
  amedaz=12
  total=120
  
• 12 years agoHelpfull: Yes(2) No(3)
• ans is 4*6=24
• 12 years agoHelpfull: Yes(2) No(4)
• 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.
• 12 years agoHelpfull: Yes(2) No(1)
• please elaborate it na unable to catch it
• 12 years agoHelpfull: Yes(0) No(0)
• friend's it should be 4!
• 12 years agoHelpfull: Yes(0) No(2)
• (AEA)taken as 1 so total letters in (AEA)ZMD are 4
  combinations possible are 4!=24.
• 12 years agoHelpfull: Yes(0) No(2)
• take AEA as a unit so total way = 4!=24
• 12 years agoHelpfull: Yes(0) No(1)
• keep aea as one set and the terms aea,m,z,d can be arranged in 4! ways...so answer is 24...
• 12 years agoHelpfull: Yes(0) No(0)