In how many ways can the digit of the number 2233558888 be arranged so that the odd digits are placed in the even positions?

• even positions=5 so the nos can be placed in 5! ways and 3 is repeated twice and also 5 is repeated twice so... 5!/2!*2!=30
  after 4 odd nos are placed then we have 6 positions for other nos to be placed so...same as 6!ways and 2 is repeated twice and 8 is repeated four times... 6!/2!*4!=15
  so 30*15=450
• 8 years agoHelpfull: Yes(27) No(0)
• odd place combination = 5!/2!*2!=30
  even place combination= 6!/2!*4!=15
  to total=30*15=450
• 8 years agoHelpfull: Yes(22) No(10)
• even no=228888, even position=5
  odd no=3355, odd position=5
  so arrangement is like{ p(5,4)/2!*2!} * {p(6,5)/2!*4!}=450
• 8 years agoHelpfull: Yes(7) No(2)
• (6!/(4!*2!)) * ((5C4*4!)/2!*2!) = 450
• 8 years agoHelpfull: Yes(4) No(2)
• There are four odd digits and five even positions. ODD digists can be placed in 4!/(2!*2!) [since 3 and 5 are repeated twice] = 6
  Again the rest of the digits can be placed in 6!/(2!*4!)=15
  Therefore no. of possible arrangements = 6*15 = 90
• 8 years agoHelpfull: Yes(3) No(3)
• 4 odd no in 5 even position =4^5/2!2!=256
  simillarly 6 even number in 5 odd position= 6^5/2!4!=162
  total way is 256*162=41472
• 8 years agoHelpfull: Yes(2) No(2)
• 5p4/(2!2!) * 6!/(4!2!) = 450
  odd even
• 8 years agoHelpfull: Yes(1) No(0)
• 5!/(2!*2!)
• 8 years agoHelpfull: Yes(0) No(0)
• 5p4*6!
  BECZ no. of even places are 5
• 8 years agoHelpfull: Yes(0) No(0)
• ans:450
  first arrange evens in odd position by:6!/2!*4!=15 ways
  and then remaining numbers arranged in =5!/2!*2!=30 ways
  total no of ways=30*15=450
  
• 7 years agoHelpfull: Yes(0) No(0)