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10 students queue up one after one randomly .Peter, Sam and Suan are in the queue and they are friends.Find the probability that they stand together in the queue one after another.

Guest Feb 11, 2015

Best Answer 

 #1
avatar+94103 
+10

The number of ways 10 people can stand in a queue is 10!

now lets treat the 3 fiends as siamese triplets.  that means there are the triplets and 7 other people in the queue.  that is 8 altogether.

that would be 8! permuations BUT  the 'triplets' can be joined in 3! orders. 

So the number of ways that the friends can be together is    8!3!

so the probability that the friends are in the queue together is      $$\frac{8!3!}{10!}=\frac{2*3}{9*10}=\frac{1*1}{3*5}=\frac{1}{15}$$

Melody  Feb 11, 2015
 #1
avatar+94103 
+10
Best Answer

The number of ways 10 people can stand in a queue is 10!

now lets treat the 3 fiends as siamese triplets.  that means there are the triplets and 7 other people in the queue.  that is 8 altogether.

that would be 8! permuations BUT  the 'triplets' can be joined in 3! orders. 

So the number of ways that the friends can be together is    8!3!

so the probability that the friends are in the queue together is      $$\frac{8!3!}{10!}=\frac{2*3}{9*10}=\frac{1*1}{3*5}=\frac{1}{15}$$

Melody  Feb 11, 2015
 #2
avatar+92622 
+5

Very nice, Melody.....maybe one day I'll catch on to probabiity.....but, it's not probable...!!!!

 

CPhill  Feb 11, 2015
 #3
avatar+94103 
0

I am sure that you have caught me once or twice in the past  :)))

And it is highly probable that you will again in the furture.  

Melody  Feb 12, 2015

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