Latest AIEEE Aptitude Question SOLUTION: There are 5 letters and 5 addressed envelopes. If the letters are put at random in the envelops, the probability that all the letters may be placed in wrongly addressed envelopes i

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(#M40021903) AIEEE Numerical Ability question PROBABILITY Keep an EYE

There are 5 letters and 5 addressed envelopes. If the letters are put at random in the envelops, the probability that all the letters may be placed in wrongly addressed envelopes is

Asked In AIEEE (10 years ago)

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GAUTAM KUMAR  (7 years ago)

I think I have found a new and simple formula to suit this particular case.

If there is one letter and one envelope then no way you can put it wrong(S1).

If there are 2 letters and 2 envelopes then you can put them wrong in 1 way(S2).

If there are 3 letters and 3 envelopes then you can put them wrong in 2 ways(S3).

If there are 4 then you can put them wrong in 9 ways(S4).

If there are 5 then you can put them wrong in 44 ways(S5).

If you observe you can find a pattern.

S3=(S1+S2)*2

S4=(S2+S3)*3

S5=(S3+S4)*4

S6=(S4+S5)*5

In general, Sn=(Sn-2 + Sn-1)*(n-1)

So, if there are 5 letters then S5=(S3+S4)*4=(2+9)*4=44

Is this solution Helpfull? Yes (18) | No (7)  

snehal daware  (9 years ago)

119 (n!-1)

Is this solution Helpfull? Yes (9) | No (42)  

pavithra venkatesan  (8 years ago)

=5!((1/2!)-(1/3!)+(1/4!)-(1/5!))
=120(1/2-1/6+1/24+1/120)
=40+5-1
=44

Is this solution Helpfull? Yes (8) | No (16)  

Karan  (6 years ago)

I have a simple technique to solve this problem:

The rule is divide n! successively with 2, 3, 4, and so on up to n. Put plus sign before even divisors and minus sign before odd divisors. Find the sum. You get the answer. For example,

Suppose, n=5;5!=120; divide 120 by 2, you get 60.

Divide 60 by 3, you get 20.

Divide 20 by 4, you get 5.

Divide 5 by 5, you get 1.

Now, do the sum: 60–20+5–1 =44.

Is this solution Helpfull? Yes (6) | No (2)  

Swasti Ranjan Senapati  (4 years ago)

From a bag containing 8 green and 5 red balls.
Therefore total balls are 8+5=13

three drawn one after the other.

the probability of all three balls beings green if the balls drawn are replace before the next ball pick = (8/13)(8/13)(8/13)
= 512/2197

the probability of all three balls drawn are not replaced (8/13)(7/12)(6/11)
= 336/1716
= 28/143

Is this solution Helpfull? Yes | No (1)  

lethaswi  (4 years ago)

my ans is 44

Is this solution Helpfull? Yes | No   

akash jaiswal  (3 years ago)

-If there is one letter and one envelope then no way you can put it wrong(S1).
If there are 2 letters and 2 envelopes then you can put them wrong in 1 way(S2).
If there are 3 letters and 3 envelopes then you can put them wrong in 2 ways(S3).
If there are 4 then you can put them wrong in 9 ways(S4).
If there are 5 then you can put them wrong in 44 ways(S5).
If you observe you can find a pattern.
S3=(S1+S2)*2
S4=(S2+S3)*3
S5=(S3+S4)*4
S6=(S4+S5)*5
In general,
Sn=(Sn-2 + Sn-1)*(n-1) So, if there are 5 letters then S5=(S3+S4)*4=(2+9)*4=44

Is this solution Helpfull? Yes | No   

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