There are 4 different letters and 4 addressed envelops. In how many ways can the letters be put in the envelops so that atleast one letter goes to the correct adress

• at least one letter goes to the correct address...
  so,
  4c4+4c3+4c2+4c1=15
  ans. 15 ways atleast 1 letter goes to the correct address
• 9 years agoHelpfull: Yes(34) No(0)
• Ans is:15
  Question is at least one of it goes to the correct address
  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..
• 9 years agoHelpfull: Yes(18) No(3)
• total no. of ways of putting 4 different letters in 4 addressed envelopes =4!=24
  no of ways when no letters gets its correct envelope =4![1-1/1!+1/2!-1/3!+1/4!]=9ways
  so, total no of ways will be =24-9=15ways
  
• 9 years agoHelpfull: Yes(8) No(0)
• ans is 15
  But many explanation is wrong.
  let envelops 1234
  and letters ABCD
  
  then number of ways 1 letter is in correct envelop
  only A is correct ADBC,ACDB
  only B is correct CBDA,DBAC
  similarly C and D
  then 8 ways
  
  number of ways 2 correct are 6
  AB is correct in ABDC
  AC is correct in ADCB
  similarly AD,BC,BD,CD
  
  number of ways 3 correct 0
  number of ways 4 correct 1
  total 8+6+1=15
  
  
  
• 9 years agoHelpfull: Yes(7) No(0)
• 15 ways.
  the number where no envelop goes to the correct address can be obtained as
  4!(1/2!-1/3!+1/4!)=9
  total ways = 4!= 24
  24-9 = 15
  
  
• 9 years agoHelpfull: Yes(4) No(0)
• 4!-non of the letters in correct address
  24-6=18
• 9 years agoHelpfull: Yes(3) No(2)
• If three letters are put into correct envelopes then automatically fourth letter will go to correct address . Thus ,
  no of ways 3 letters are correct=no. of ways all four are correct= 4c4=1
  therefore,total no of ways are 4c1+4c2+4c3=11.
• 9 years agoHelpfull: Yes(2) No(1)
• One letter goes correct-6 ways
  Two letter goes correct-2 ways
  Three letter -1
  Four letter-1
  Total ten ways
• 9 years agoHelpfull: Yes(1) No(6)
• 7 should be the answer...
  1 letter correct- 3!-1=5(i did -1 because when we are taking 3! All correct case gets included in it which is 1 way...so we need to substract that)
  2 letter correct- 2!-1=1
  3=4 letter correct- 1
  Hence 7
• 9 years agoHelpfull: Yes(1) No(1)
• 9 ways only bcoz for 3 lettes at the right place is same as all the letters at the right place..
  1 letters rite=3!=6
  2 letters rite=2!=4
  3 & 4 letters rite=1 ways
  so total=9
• 9 years agoHelpfull: Yes(0) No(1)
• one letter - 4 ways
  two letter- 6 ways
  3 letter- 1 ways
  4 letter - 1 ways
  total 4+6+1+1= 12
• 9 years agoHelpfull: Yes(0) No(0)
• actually 3 and 4 letter goes to the correct address only 1 way because if 3 letter goes to the correct address 4th letter automatically goes to the correct.
  so,ans will be 1+4c2+4c1=11
• 9 years agoHelpfull: Yes(0) No(0)
• 4c4+4c1+4c2=1+4+6 only 11 possibles are there because if one letter is in correct there should at least one more letter have mis match address
  
  so ans:- 11
• 9 years agoHelpfull: Yes(0) No(0)