what is probabability that 5 envelope can not put in to the right envelope?

• Number of derangements = n! (1/2! - 1/3! + 1/4! + ... + ((-1)^n)/n!)
  
  Given 5 letters A, B, C, D and E:
  Total number of derangements = 5! (1/2! - 1/3! + 1/4! - 1/5!) = 60-20+5-1 = 44.
  Total possible arrangements = 5! = 120.
  P(no letter is in the correct position) = 44/120 = 11/30.
• 9 years agoHelpfull: Yes(29) No(1)
• answer is 4^5/5^5
  
• 9 years agoHelpfull: Yes(25) No(15)
• everybody listen,
  
  Mr.Dhruv's answer is right please follow that, i have verified a lot about this topic.
  
  5! (1/2! - 1/3! + 1/4! - 1/5!) = 60-20+5-1 = 44
  
  answer : 44/120
  
• 9 years agoHelpfull: Yes(8) No(0)
• ans is zero
• 9 years agoHelpfull: Yes(5) No(9)
• 1 every envelope is not getting in right place dat means 1/5+1/5+1/5+1/5+1/5
• 9 years agoHelpfull: Yes(1) No(5)
• Total no of ways of distributing 5 letters in 5 envelope is 5! or 720 ways....and there is only one way for distributing 5 letter in 5 right envelope... Hence remaining wrong ways are 720-1=719
  Hence probability will be 719720
• 9 years agoHelpfull: Yes(1) No(3)
• 36
  5!(1/0!-1/1!+1/2!-1/3!+1/4!-1/5!)
  
• 9 years agoHelpfull: Yes(0) No(3)
• ans is zero ......since it is not possible for all 5 to be put in wrong envelope.....if 4 of them goes in wrong then 1 definitely has to go to the right envelope
• 9 years agoHelpfull: Yes(0) No(1)
• Sample space5!
  Event is (5!-1)/5!
  1is the no of ways of placing five letters correctly so 1-(1/5!)is the ans
• 9 years agoHelpfull: Yes(0) No(2)
• Since for letter to b placed in correct envelope we have 1/5!
  Letters are not placed in correct envelope 1-1/5!
  Required probability 119/120 refer to arun sharma probability lod 3
• 9 years agoHelpfull: Yes(0) No(0)