There are 4 different letters and 4 addressed envelopes.In how many ways can the letters be put in the envelopes so that atleast one letter goes to the correct address ?
a)15
b)16
c)18
d)12

• it is given that atleast one of it goes to correct address ie.there is apossiblity that 2 goes correct,3 goes correct,4 goes correct.
  therefore no.of ways 1 goes correct is 4c1=4
  no. of ways 2 goes correct is 4c2=6
  no. of ways 3 goes correct is 4c3=4
  no. of ways 4 goes correct is 4c4=1
  therefore total no. of ways is 4+6+4+1=15(ans)
• 10 years agoHelpfull: Yes(119) No(18)
• no.of ways to put all letters correctly=1
  ways to put 2 letters correctly=4C2=6
  ways to put 1 letter correctly=2*2*2=8
  total=1+6+8=15
• 10 years agoHelpfull: Yes(18) No(9)
• For each envelope there are two possibilities in terms of holding a letter:
  1)Holding a correct letter 2)Incorrect letter.
  So in terms of correctness of letters in envelops there are 2*2*2*2 = 16 possible cases.
  But among them one case is that all the letters are wrongly placed. So subtract that one possibility.
  Thus the ans is 16 - 1 = 15 ways
• 10 years agoHelpfull: Yes(16) No(2)
• Total ways of putting r letters to r covers = r! = 4! = 24
  Number of ways that none of them goes into the right envolope = D 4 =4!(1 2! −1 3! +1 4! ) = 9
  So atleast one envolope goes into the right one = 24 - 9 = 15
  
• 10 years agoHelpfull: Yes(6) No(5)
• 4c1+4c2+4c3+4c4= 15
• 10 years agoHelpfull: Yes(5) No(1)
• 12 ways.4c2* 2!=12
  
• 10 years agoHelpfull: Yes(4) No(12)
• It CAN BE SOLVE BY TAKING HELP OF DEARRANGEMENTS
  
  No. of possible arrangements so that out of n objects none goes to its corr BE ect/original position is D(n)
  D(n)= n! (1 - 1/1! + 1/2! - 1/3! + ... +(-1)^n 1/n! )
  
  D(4)=4!(1-1/1!+1/2!-1/3!+1/4!)
  =24(1-1+1/2-1/6+1/24)
  =12-4+1=9
  
  
  ramaining may atleast one,or two or three or four or all goes to correct position
  
  total arrangements-no goes to correct
  24-9=15
• 8 years agoHelpfull: Yes(4) No(0)
• @SUBARNA SEN...correct answer is a)15
• 10 years agoHelpfull: Yes(3) No(5)
• t is given that atleast one of it goes to correct address ie.there is apossiblity that 2 goes correct,3 goes correct,4 goes correct.
  if 3 letter goes correct one letter will go to any invalid address,if that one letter goes invalid address then another letter must go to wrong address,so we will discard (3 letter goes correct situation).
  so,therefore no.of ways 1 goes correct is 4c1=4
  no. of ways 2 goes correct is 4c2=6
  
  no. of ways 4 goes correct is 4c4=1
  therefore total no. of ways is 4+6+1=11(ans)
  
• 9 years agoHelpfull: Yes(3) No(3)
• ans is 15
• 10 years agoHelpfull: Yes(1) No(0)
• @Ambuj Srivastav :plz xplain
  
• 10 years agoHelpfull: Yes(0) No(0)
• 4c1+4c2+4c3+4c4=15
• 7 years agoHelpfull: Yes(0) No(0)