the letters in the roadie are permuted in all possible ways and arranged in alphabetical order.find the word in the 44th rank?

• AERDOI
  having AD as first 2 letters we can have 24 words. then AED we have 6 words, ie 24+6=30.
  having AEI again 6 words, ie 30+6= 36, having AEO, 36+6=42. AERD we have 2 words, AERDIO and AERDOI. since in alphabetical order, answer is AERDOI
• 9 years agoHelpfull: Yes(19) No(1)
• A[DEIOR] NOT POSSIBLE AS 5! IS MUCH HIGH THAN OUR NEED .
  AD[EIOR] = 4! = 24
  AE[DIOR] = 4! = 24
  AED[1OR] = 3! = 6+24=30
  AEI[DOR] =3! = 6+30 =36
  AEO[DIR] = 3! = 36+6 =42
  NOW 43RD LETTER WOULD BE AERDIO
  AND HENCE 44TH IS "AERDOI" -ANSWER .
• 9 years agoHelpfull: Yes(14) No(0)
• A(DEIOR)120 IS NOT POSSIBLE
  AD(EIOR) = 4! = 24
  AE[DIOR] = 4! = 24
  AED[1OR] = 3! = 6+24=30
  AEI[DOR] =3! = 6+30 =36
  AEO[DIR] = 3! = 36+6 =42
  AERDOI IS 44TH WORD
• 9 years agoHelpfull: Yes(4) No(1)
• PLEASE EXPLAIN MORE CLEARLY HOW TO SOLVE THIS TYPE OF QUESTIONS FIRST WE CONSIDER AD THEN WHY DO WE GO FOR AEI THOUGH IT IS NOT IN ALPHABETICAL ORDER
• 9 years agoHelpfull: Yes(2) No(1)
• After arranging the letters we get ADEIOR
  now,
  AD (4!) = 24
  AED (3!) = 6
  AEI (3!) = 6
  AEO (3!) = 6
  AERD (2!) = 2
  So total 44.
  Now 43rd word is AERDIO
  and 44th word is "AERDOI"
  
• 9 years agoHelpfull: Yes(2) No(1)
• AERDOI
  BCOZ I SOLVED IT
• 9 years agoHelpfull: Yes(1) No(0)
• AERDOI IS THE RIGHT ANSWER
• 9 years agoHelpfull: Yes(1) No(0)
• letters starting with AE= 4!=24
  then letters starting with ADE= 3!=6
  then Letters starting with ADI=3!=6
  then letters starting with ADO= 3!=6
  
  .
  upto now, we have first 42 words of the dictionary..
  then the next word is ADOEIR which is the 3rd word, then is ADOERI which is the 44th word.
• 9 years agoHelpfull: Yes(0) No(9)
• plz explain how to solve these type of qstns...?
• 9 years agoHelpfull: Yes(0) No(1)
• adeior [in alphabetic order]
  adei[or]=2!= 2
  ade[ior]=3!= 6
  ad[eior]=4!= 24 [5! is 120 so not possible] { 2+6+24=32, the remaining 12 combinations can be obtained as follows}
  aed[ior]=3!= 6
  aei[dor]=3!= 6 [24+6+6=44]
  so now in the last 6 combinations
  dor,
  dro,
  odr,
  ord,
  rdo,
  rod=44th word
  therefore the 44th rank is AEIROD
  
• 9 years agoHelpfull: Yes(0) No(1)
• 44th rnk is AERDOI
• 8 years agoHelpfull: Yes(0) No(0)