Latest TCS Aptitude Question SOLUTION: 5 letters and 5 addresses are there then what is the probability that no single letter goes to the correct address......try and tell me answer with clear explanation....

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(#M40154930) TCS Numerical Ability question Permutation and Combinations , Probability Keep an EYE

5 letters and 5 addresses are there then what is the probability that no single letter goes to the correct address......try and tell me answer with clear explanation....

Asked In TCS N Murali Krishna (8 years ago)

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Mukesh  (8 years ago)

above formula is : n!(1/2! - 1/3! + 1/4!+ .......((-1)^n)/n! by putting : into this formula : 5! -76 /5! = 11/30

Is this solution Helpfull? Yes (10) | No (2)  

Kiran Choudhary  (8 years ago)

Total No of ways in which at least one letter goes to the correct envelope is 5!(1-1/2!+1/3!-1/4!+1/5!)=76So the no of ways in which none of the letter goes into the correct envelope = 5! - 76=44So reqd probability = 44/120 =11/30

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abhishek anand  (8 years ago)

here 1st letter can be put in any one of the 5 envelopes in 5 ways. 2nd letter can be put in any one of the 4 envelopes in 4 ways and so on then we have 5*4*3*2*1=120
out of 120 only one is the correct way to do so therefore total incorrect ways=120-1=119.

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Manjunath  (8 years ago)

probabilty of no one go to currect place=1-all will go to currect place
=1-(1/5*1/4*1/3*1/2*1/1)=1/120

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Mukesh  (8 years ago)

Question goes like this : Five letters are addressed to five different persons and the corresponding envelopes are prepared. The letters are put into the envelopes at random. What is the probability that no letter is in its proper envelope?

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Mukesh  (8 years ago)

So let the letters be : A B C D E and the corresponding envelopes are : a b c d e. Now A cannot go to a, so A can for example go to c. Then C has 4 options to go to: a , b, d, and e. The 1st has 2 choices, the 2nd has 3, the 3rd has 3, and the 4th has 3 choices. So there are 2 + 3 + 3 + 3 = 11 choices. Since A has 4 options to go to, the total is then : 11*4 = 44. And there are a total of 5! ways to do so. So then probability is : 44/120. This is perhaps easier to digest than using the full theorem.

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Mukesh  (8 years ago)

Or we can directly put this a formula : The formula for the derangement can be used directly. The number of derangements is
n!(12!−13!+14!+…+(−1)nn!)
n!(12!−13!+14!+…+(−1)nn!)
put the value of n and get your answer.

i guess this is the fastest and safest way. hope it helpful. cheers.

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gopinath  (8 years ago)

Hi friends. Is anybody here who got hall ticket and taken the tcs test today except karnataka state candidates.. Please tell me the difficulty level . :)

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vivek kumar  (8 years ago)

5 letters can be goes in different adress in 5! way and there is only condition in which it goes to proper adress so ans is 5!-1=119

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MANISH DHYANI  (7 years ago)

correct answer is 4^5/5^5=1024/3125

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bhumika  (4 years ago)

Dearr(5) = 5! (1/0! - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!)
Dearr(5) = 120 (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
Dearr(5) = 60 - 20 + 5 - 1 = 44

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