Latest Elitmus Aptitude Question SOLUTION: There are 5 envelopes corresponding to 5 letters.If the letters are placed in the envelopes at random,what is the probability that no single letter is placed in the right envelope?

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(#M40152090) ELITMUS Numerical Ability question Probability Keep an EYE

There are 5 envelopes corresponding to 5 letters.If the letters are placed in the envelopes at random,what is the probability that no single letter is placed in the right envelope?

Asked In Elitmus Souradeep Guhathakurta (8 years ago)

Solved kankana sarkar Read Solution (18)

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Read Solution (18) :

kankana sarkar  (8 years ago) Best Solution

this is the disarrangement problem.. so no of ways all 5 letter will go in wrong place= 5!*[1/2!-1/3!+1/4!-1/5!]=60-20+5-1=44. total case= 5!=120.
so probability that no one is in right envelope=44/120

Is this solution Helpfull? Yes (55) | No (8)  

DEEPMALA TIWARI  (8 years ago)

119/120
5 letters are placed in 5 eneveopes in 5! ways
all letters are in correct envelopes prob. 1/5!
so prob. of not in correct enevelpoe 1-1/5!

Is this solution Helpfull? Yes (9) | No (16)  

abhilash chhipa  (8 years ago)

total possibilities= 5!
only one can be correct out of these 5!
so the required probability will be (5!-1)/5!=119/120

Is this solution Helpfull? Yes (5) | No (7)  

Abhishek nkumar  (8 years ago)

ans. should be 0.

Is this solution Helpfull? Yes (4) | No (28)  

Ranjita Rani Naik  (8 years ago)

ans.44/120

Is this solution Helpfull? Yes (3) | No (6)  

jithin prakash  (8 years ago)

no of ways=5!=120
only one arrangement is right position.
so probability for right position is 1/120(for one letter)
probability for 5 letter is 5/120=1/24
so probability for wrong place=1-1/24=23/24

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

sarath  (8 years ago)

5 envelopes corresponding to 5 letters i.e in an random way possible combinations is 5c5,
the possibility of all letters is placed in right envelope is 5c5
one thing we have to remember in probability is p(s)=1 (ALWAYS)
that y the possibility for not getting even single letter can be given as
(p(s)-getting all letter placed in right envelope)
=1-5c5
=1-1
=0

Is this solution Helpfull? Yes (2) | No (9)  

saikumar  (8 years ago)

4*3*2/120=
1/5.....

Is this solution Helpfull? Yes (2) | No (5)  

JUGAL KHAJURIYA  (8 years ago)

ans is 119/120

Is this solution Helpfull? Yes (1) | No (4)  

RITESH DEO  (8 years ago)

Answer is 0

because... probability that all letter is placed in right envelpo = 5C5
now probability that none of the lettr is placed in the right envelop= 1- 5C5 = 1-1= 0

Is this solution Helpfull? Yes (1) | No (2)  

SAURAV  (7 years ago)

The total number of solution of disarrangement problem can be solved as :
No of letters(n) = 5
D(n)=n![1/1!-1/2!+1/3!-1/4!+.....(-1)n /n!]
W=n!
solution=D(n)/W

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

pulugam geetha  (8 years ago)

there is only one way to place in the right envelope
but 5 lettrers are placed in 5 envelopes in 5!ways
so probability of not in correct enevelope is 5! -1=119

Is this solution Helpfull? Yes | No (3)  

niraj kumar  (8 years ago)

prob for going in correct envelope :- 1/5
so for going in wrong envelope :- 1-1/5= 4/5

Is this solution Helpfull? Yes | No (6)  

lokesh kumar  (8 years ago)

how 19????????/

Is this solution Helpfull? Yes | No (1)  

Aniruddha  (8 years ago)

Answer is zero as it is impossible to put all letters wrong

Is this solution Helpfull? Yes | No (7)  

Chitaranjan kumar  (8 years ago)

4*4*4*4*4=1024
bcz except fixed one others 4 çan put for every 5.

Is this solution Helpfull? Yes | No (1)  

srishti snehi  (5 years ago)

it's a question of dearrangement .ans = 26 .

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Asif Hisain Jafri  (4 years ago)

e are 5 letters and five addressed envelopes. the number of ways in which all the letters can be put in wrong envelopes is

Total Number of ways = 5! = 120

The formula for the derangement can be used directly. The number of derangements is

n! (1/2!−1/3!+1/4!+…+(−1)ⁿ/n!)

Here n = 5

5 ! ( 1/2! - 1/3! + 1/4! - 1/5!)

= 120 ( 1/2 - 1/6 + 1/24 - 1/120)

= 60 - 20 + 5 - 1

= 65 - 21

= 44
So probabilty = 44/120

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