Latest Elitmus Aptitude Question SOLUTION: Raju and Reenu are standing in a row there are 9 peoples. what is the probability that at least 3 people will stand between Raju and Reenu

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(#M40152365) ELITMUS Numerical Ability question Probability Keep an EYE

Raju and Reenu are standing in a row there are 9 peoples. what is the probability that at least 3 people will stand between Raju and Reenu

Asked In Elitmus Soumen Maity (9 years ago)

Solved Ananya Maiti Read Solution (9)

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Ananya Maiti  (9 years ago) Best Solution

there are at least 3 people between Raju and Reena... people can be 4,5,6,7
when there are 3 people between them... in 5*7!*2!
when there are 4 people between them... in 4*7!*2!
when there are 5 people between them... in 3*7!*2!
when there are 6 people between them... in 2*7!*2!
when there are 7 people between them... in 1*7!*2!
so total 15*7!*2!....
total no of possibilities... 9!
probability will be 15*7!*2!/9!=5/12

Is this solution Helpfull? Yes (12) | No (6)  

RAHUL SAHA  (9 years ago)

9C2=36
3 people between Raju and Reena 7+5+3=15
15/36=5/12 Ans

Is this solution Helpfull? Yes (10) | No (3)  

Abhishek nkumar  (9 years ago)

A shopkeeper has three types of rice pocket.all pocket is multiple of one kg. there are 4 persons A, B,C,D, each take exact two packet. and take 11kg,12kg,16kg, and 17kg respectively then

Q1.how much maximum weight taken by any person

Q2. how much minimum weight taken by any person
ASKED in yesterday elitmus Exam

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Mukesh  (9 years ago)

11 = 6 + 5
12 = 6 + 6
16 = 5 + 11
17 = 6 + 11
So, Max Weight = 11 + 11 = 22
And Min Weight = 5 + 5 =10

Is this solution Helpfull? Yes (7) | No (3)  

Pankaj Kumar  (9 years ago)

when 3 people b/w them. then - 5*7!*2!
when 4 people b/w them. then- 4*7!*2!
when 5 people b/w them.then- 3*7!*2!
when 6 people b/w them.then- 2*7!*2!
when 7 people b/w them.then- 1*7!*2!
total no of possibilities... 9!
then Ans will be 15*7!*2!/9!=5/12

Is this solution Helpfull? Yes (4) | No   

vikash  (6 years ago)

link ---
https://math.stackexchange.com/questions/1397099/what-is-the-probability-that-raj-and-rana-have-atleast-3-persons-between-them-gi

Is this solution Helpfull? Yes (2) | No   

Kartik Choudhary  (5 years ago)

9 people can be arranged in 9! ways. Excluding Raj and Rana, out of 7 people.

Case 1: I select 3 people 7C3 who can be arranged in 3! ways.

Remaining (9−(3+2))=4 people can be arranged in 4! ways

Therefore required arrangement=7C3∗3!∗4!∗2! ----(i)
Case 2: I select 4 people 7C4 who can be arranged in 4! ways

Remaining (9−(4+2))=3 people can be arranged in 3! ways

Therefore required arrangement=7C4∗4!∗3!∗2! ----(ii)
Case 3: I select 5 people 7C5 who can be arranged in 5! ways

Remaining (9−(5+2))=2 people can be arranged in 2! ways

Therefore required arrangement=7C3∗5!∗2!∗2! ----(iii)
Case 4: I select 6 people 7C6 who can be arranged in 6! ways

Remaining (9−(6+2))=1 people can -be arranged in 1! way.

Therefore required arrangement=7C6∗6!∗1!∗2! ----(iv)
Case 5: I select 7 people 7C7 which can be arranged in 7! ways

Remaining (9−(7+2))=0 people can be arranged in 1! ways

Therefore required arrangement=7C7∗7!∗1!∗2! ----(v)
Adding equations (i)+(ii)+(iii)+(iv)+(v)9! we get the required probability.

Is it a correct approach and if there exists any shorter method, could you please tell me?

Correct answer is 512.

Is this solution Helpfull? Yes (1) | No   

Arpit dubey  (9 years ago)

in this question we didnt need to care about the positioning of raju n reena ryt??? like whether raju is placed first or reena?

Is this solution Helpfull? Yes | No (1)  

insha singh  (8 years ago)

what does 5 , 4 , 3... signify in the ans. why arent we using 5! , 4!...?

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