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.
+118609 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 $$\frac{8!3!}{10!}=\frac{2*3}{9*10}=\frac{1*1}{3*5}=\frac{1}{15}$$
+118609 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 $$\frac{8!3!}{10!}=\frac{2*3}{9*10}=\frac{1*1}{3*5}=\frac{1}{15}$$
