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Updated on: 10 Mar 2016, 00:23
Originally posted by vardhanindaram on 10 Mar 2016, 00:19.
Last edited by Bunuel on 10 Mar 2016, 00:23, edited 1 time in total.
Last edited by Bunuel on 10 Mar 2016, 00:23, edited 1 time in total.
Renamed the topic and edited the question.
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
40% (02:14) correct
60%
(02:34)
wrong
based on 156
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5 letters have to be put in 5 different envelopes numbered 1 through 5 such that each of the letters go into only 1 envelope.The letter is said to be put in correct position if for example letter 1 goes into envelope 1.Now what is the probability that all letters be put into wrong envelopes?
A. 1/3
B. 2/3
C. 11/120
D. 11/30
E. 76/120
A. 1/3
B. 2/3
C. 11/120
D. 11/30
E. 76/120
10 Mar 2016, 01:15
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vardhanindaram
This Q is based on derangement, a permutation in which all elements are in the wrong position.
Number of derangements = \(n! (\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} + ... + (\frac{(-1)^n)}{n!})\).. Ofcourse it can be derived but thats not required..
Since there are 5 letters and 5 envelopes:-
\(Derangements = 5! (\frac{1}{2!} - \frac{1}{3!}+ \frac{1}{4!}- \frac{1}{5!})\)
=> \(120(\frac{1}{2} - \frac{1}{6}+\frac{1}{24}- \frac{1}{120}) = 60 -20 +5 -1=44.\)
Total possible ways = 5! = 120.
therefore required prob, \(P = \frac{44}{120} = \frac{11}{30}.\)
ans D
AR15J
Putting the letter L3 in E1 OR in E2/E4/E5 are different cases
If L3 is put in E1, we have 3 letters and 3 envelopes - L2, L4, L5 and E2, E4, E5
All three need to be put in different envelopes. There are 2 ways of doing this.
In this case, there are 4*1*2 = 8 ways of putting all the letters incorrectly.
The rest of your solution goes like this now:
Or L3 can be put in E2/E4/E5 in 3 ways. Let's say it is put in E2.
now, L2 can be put incorrectly in E1, E4 or E5. Again, now if L2 is put in E1, there is only 1 way to put L4 and L5 incorrectly.
But if L2 is put in E4 (or E5), then L4 (or L5) can be put in 2 ways.
So there are in all 3 ways to put L2 and the remaining letters.
so, total combinations are 4*3*3=36
Required Probability = (8 + 36)/120 = 44/120 = 11/30
