Latest TCS Aptitude Question SOLUTION: 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

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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

Asked In TCS ranjan kumar (9 years ago)

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Sourick  (9 years ago)

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

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karthikeyank  (9 years ago)

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..

Is this solution Helpfull? Yes (18) | No (3)  

RUPESH KUMAR  (9 years ago)

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

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Parshant garg  (9 years ago)

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

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venkatesh  (9 years ago)

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

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Ajay Singh Yadav  (9 years ago)

4!-non of the letters in correct address
24-6=18

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Harneet janda  (9 years ago)

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.

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tirupathi reddy  (9 years ago)

One letter goes correct-6 ways
Two letter goes correct-2 ways
Three letter -1
Four letter-1
Total ten ways

Is this solution Helpfull? Yes (1) | No (6)  

Swayanti Bhaumik  (9 years ago)

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

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GAURAV KUMAR  (9 years ago)

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

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Ria Ghosh  (9 years ago)

one letter - 4 ways
two letter- 6 ways
3 letter- 1 ways
4 letter - 1 ways
total 4+6+1+1= 12

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Sourick  (9 years ago)

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

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csemalli  (9 years ago)

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

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