Latest TCS Aptitude Question SOLUTION: 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

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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

Asked In TCS Ambuj Shrivastava (10 years ago)

Solved sandy Read Solution (12)

Is this Puzzle helpful?   (48)   (17) Submit Your Solution Permutation and Combination

Read Solution (12) :

sandy  (10 years ago)

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)

Is this solution Helpfull? Yes (119) | No (18)  

bhaavika panjwani  (10 years ago)

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

Is this solution Helpfull? Yes (18) | No (9)  

Inty Jagannath Suhit  (10 years ago)

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

Is this solution Helpfull? Yes (16) | No (2)  

aditi patel  (10 years ago)

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

Is this solution Helpfull? Yes (6) | No (5)  

Ambuj Shrivastava  (10 years ago)

4c1+4c2+4c3+4c4= 15

Is this solution Helpfull? Yes (5) | No (1)  

subarna sen  (10 years ago)

12 ways.4c2* 2!=12

Is this solution Helpfull? Yes (4) | No (12)  

Vandan Mishra  (7 years ago)

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

Is this solution Helpfull? Yes (4) | No   

Ambuj Shrivastava  (10 years ago)

@SUBARNA SEN...correct answer is a)15

Is this solution Helpfull? Yes (3) | No (5)  

aninda kar  (9 years ago)

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)

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

ankita gajanan kulkarni  (10 years ago)

ans is 15

Is this solution Helpfull? Yes (1) | No   

subarna sen  (10 years ago)

@Ambuj Srivastav :plz xplain

Is this solution Helpfull? Yes | No   

garima agrawal  (7 years ago)

4c1+4c2+4c3+4c4=15

Is this solution Helpfull? Yes | No   

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