What is the probability that when 4 letters are placed in 4 addressed envelope,all r wrongly placed?

• the ans is: 3/8
  
  Total no. of ways to put the letters in the envelop is !4= 24
  
  all letters are wrongly placed..
  suppose the letters are 1,2,3,4
  if i write 1234 it means in envelop 1 there is 1st letter, in 2 there is 2nd letter ..so on
  
  there would be 9 such combination to put the letters in the wrong envelop:
  2143, 2341, 2413
  3142, 3412, 3421
  4123, 4312, 4321
  
  so 9/24= 3/8 enjoy :)
• 13 years agoHelpfull: Yes(14) No(3)
• there is one formula
  n!(1/2!-1/3!+1/4!)
  4!(1/2!-1/3!+1/4!)
  24*9/24=9 is the ans...
  
• 13 years agoHelpfull: Yes(6) No(5)
• All Are Wrongly Placed(Or) None Is Correctly Placed Both Is Same Na...So Probability Will Be (1/2!)-(1/3!)+((1/4!)=3/8
• 13 years agoHelpfull: Yes(5) No(0)
• it is a derangement question. for which formula is 4!(1-1/1!+1/2!-1/3!+1/4!)=9
• 13 years agoHelpfull: Yes(3) No(4)
• probability of keeping 4 letters is 2^4=16
  in dat, crct way is 1.
  so, wrong ways are 2^4-1=15
• 13 years agoHelpfull: Yes(1) No(9)
• prob can never be greater than 1.
  prob that letter is put in correct envelope is 1/4.
  so ans is 1-1/4=3/4
• 13 years agoHelpfull: Yes(1) No(3)
• nCr=n!/r!(n-r)!
  Hence 4C4=12/12(1)=1 where 0!=1
  
  So the probability is 1.
• 13 years agoHelpfull: Yes(0) No(3)
• probability can never be greater than 1 then how it could be 16 or 9.
  total outcomes=16
  favourable outcomes=1(becouse there can only be 1 situation in which all the 4 letters are wrongly placed).
  so the probability=1/16.
• 13 years agoHelpfull: Yes(0) No(3)