Latest Elitmus Aptitude Question SOLUTION: There are 9 players including Mic and Jordan standing in a row. What is the probability of being 2 or less players between Mic and Jordan.

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(#M40019680) ELITMUS Numerical Ability question PROBABILITY Keep an EYE

There are 9 players including Mic and Jordan standing in a row. What is the probability of being 2 or less players between Mic and Jordan.

Asked In Elitmus Ratnesh Pandey (10 years ago)

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Read Solution (8) :

Mohit Awasthi  (10 years ago)

total no. of ways arranging 9 people is = 9factorial.
case-1--
both sit together
hence no. of ways of arranging them is 8fact*2fact.
case-2
1person sit between them
hence no. of way of arranging them is 7(2fact*7fact).
case--3
2 person sit between them
hence no. of ways arranging them is 7(2fact*6fact)
hence total no. of ways ==8!2!+7(2!7!)+6(2!7!)
divide this by 9!..
ans.. 7/12

Is this solution Helpfull? Yes (23) | No (5)  

Narendra Pargain  (10 years ago)

7/12....
Simple one...

Is this solution Helpfull? Yes (17) | No (8)  

Ratnesh Pandey  (10 years ago)

@prerana
the events of arranging 2 players in a row of 9= 9c2= 36
the events when there will be 2 or less players between Mic and Jordan= 21
so the probability of being 2 or less players between Mic and Jordan= 21/36 = 7/12

Is this solution Helpfull? Yes (5) | No (8)  

Gaurav Malhotra  (9 years ago)

@MOHIT AWASTHI the case whn 2 persons sit btw looks to be wrng...

it shd be 7p2(2fact*6fact)

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pooja  (10 years ago)

explain it properly..plz

Is this solution Helpfull? Yes (1) | No (7)  

prerana  (10 years ago)

please explain

Is this solution Helpfull? Yes | No (7)  

tanya  (7 years ago)

why are we multiplying by 7 in case 2 and 3 i.e in 7(2fact*7fact) ?? please explain

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MOHAMMAD ANSAB  (7 years ago)

In case 2 :
Both Mic and Jordan with the 1person standing in between is taken as a combination and the others as 6 different arrangements.so total of 7.
this can be done in 7 different ways.so multipying it with 7

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