How many different 9 digit numbers can be formed from the number 223355888 by re-arranging its digits so that the odd digits occupy even position?

• total number of ways are 60
  
  there are 5 even numbers which have to be placed at odd places
  in these numbers 2 repeats two times and 8 repeats three times
  so number of arranging these 5 even numbers are 5!/(2!3!) = 10 ways
  
  there are 4 odd numbers which have to be placed at even positions
  in these 4 odd numbers 3 and 5 repeat two times
  so the ways of arranging these 4 numbers are 4!/(2!2!) = 6 ways
  
  so total number of ways of arranging all these numbers are 10 * 6 = 60 ways
• 11 years agoHelpfull: Yes(44) No(5)
• 4!*5!/(2!*2!*2!*3!)
• 11 years agoHelpfull: Yes(5) No(0)
• 4!*5!/2!*2!*2!*3!
  60 ways
• 11 years agoHelpfull: Yes(3) No(0)
• answer would be 60
• 11 years agoHelpfull: Yes(1) No(0)
• Let one arrangement be 232385858.
  Now permute digits at odd places= 5!/(2!*3!)=10
  Now permute digits at even places= 4!/(2!*2!)=6
  
  Therefore total ways= 10*6=60.
• 11 years agoHelpfull: Yes(1) No(1)
• No.of possibilities for 4 odd numbers are in 5 odd positions so 5!/2!*2!
  No.of ways for 6 even numbers are 6!/2!*4!
  Multiplication of these two solutions is 450 ways
  
• 9 years agoHelpfull: Yes(1) No(1)
• ANS = 450
  (5C4 x 4! / 2!x2!) x ( 6C6 x 6! / 2!x4!)
• 9 years agoHelpfull: Yes(1) No(1)
• 223355888
  odd place=5!/(2!*2!)=30
  even place=6!/(2!*4!)=15
  30*15=450
• 9 years agoHelpfull: Yes(1) No(1)
• ans is 60.
• 9 years agoHelpfull: Yes(1) No(1)
• 120*5 =720
• 11 years agoHelpfull: Yes(0) No(7)
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• 11 years agoHelpfull: Yes(0) No(3)
• If there are 10 digits like this 2233558888, what will be the solution?
• 10 years agoHelpfull: Yes(0) No(2)
• reply for saravanan
  
  no of odd digits=4
  no of even digitd-6
  no of even positions=5
  then 5p4/2!2!*6!
• 10 years agoHelpfull: Yes(0) No(1)