The letters of the word AMIT when arranged as per dictionary, find the 12th word.

• ITMA
  a----=6
  IA--=2(8 WORDS)
  IM--=2(10 WORDS)
  ITAM IS 11TH WORD
  ITMA IS 12TH WORD
• 9 years agoHelpfull: Yes(29) No(3)
• Dictionary word is = AIMT
  Now A[ IMT ] = 3! = 6
  Next word will be IAMT
  I[ AMT ] = 6 + 3! = 12
  Now the last 3 words [ AMT ] are fully arranged so they became [ TMA ]
  So our ans will be I[ TMA ] = ITMA
• 9 years agoHelpfull: Yes(9) No(0)
• arranging them in alphabitcly we get AIMT... now with 1st letter a should get 6 word.. we hav to get 12th word.. so now come to the words starting with IA -2..(total is now 6+2=8).. now come to the words starting with IM-2(total is now 8+2=10).. now if we find the word starting with IT we will get the 12th word(10+2=12).. so the two consecutive words starting with IT is ITAM and ITMA.. so the ans should b ITMA
• 9 years agoHelpfull: Yes(2) No(0)
• 3210
  MAIT
  no. of letters smaller then M in RHS are 2,
  no. of letters smaller then A in RHS are 0,
  no. of letters smaller then I in RHS are 0,
  no. of letters smaller then T in RHS are 0,
  So 3!*2+0+0+0+0=12
  So MAITis 12th word.
  
  
• 9 years agoHelpfull: Yes(1) No(8)
• can any1 say what the exact way to solve this type of questions...plz!!!
• 9 years agoHelpfull: Yes(1) No(0)
• AMIT
  A[IMT]=3! =6
  AI[MT]=2!=2
  AM[MT]=2!=2
  AT[IM]=2!=2
  SO the word at 12th position is ATIM
• 9 years agoHelpfull: Yes(1) No(2)
• i cant understand please clarify in simple manner.
  
• 9 years agoHelpfull: Yes(0) No(2)
• can any one tell how to solve this type of prblm...
  
• 9 years agoHelpfull: Yes(0) No(0)
• AMIT
  in alphabetical order AIMT
  total words can be made 4!
  A - - - for these three 3!=6
  I - - - for these three 3!=6
  The last word, start with I is ITMA
• 9 years agoHelpfull: Yes(0) No(0)
• given AMIT
  Start A*3!=6
  if start next order i.e I*3!=6
  so,6+6=12
  so,answer =ITMA
• 9 years agoHelpfull: Yes(0) No(0)
• how the words are arranged some one please post the first few words
• 9 years agoHelpfull: Yes(0) No(0)