mehrasa wrote:Eric and Ortega and their teammates watch a movie. They all sit in a row, and they can sit in n
different ways. In how many of the ways can Eric sit to the right of Ortega?
(A) n
(B) n/2
(C) n/3
(D) n/4
(E) n/5

It is equally likely that Eric sits to the right of Ortega or not. Therefore, out of n possible arrangements, in exactly half of n arrangements, Eric sits to the right of Ortega, which is n/2.

The correct answer is B.

mehrasa wrote:how you are saying that "It is equally likely that Eric sits to the right of Ortega or not". since we do not know about the number of people sitting in the row,the place of eric and ortega can vary. eric can sit to the right of ortega, left of ortega or even in another place where there are other people between... need more clarification.. thnx

It is equally likely because there are only 2 possibilities, irrespective of how many friends are there:
(1) Eric can either sit to the right of Ortega
Or
(2) Eric can sit to the left of Ortega.

Two or more events are said to be equally likely, if on taking into account all the given conditions, there is no reason to accept any one of the events in preference over the others.

mehrasa wrote:I got the point.. thank you for your further explaination

You are welcome!

mehrasa wrote:Eric and Ortega and their teammates watch a movie. They all sit in a row, and they can sit in n
different ways. In how many of the ways can Eric sit to the right of Ortega?
(A) n
(B) n/2
(C) n/3
(D) n/4
(E) n/5

We should notice that for any arrangement where Eric sits to the left of Ortega, if we switch the positions of Eric and Ortega and keep everyone else fixed, we will obtain an arrangement in which Eric sits to the right of Ortega. Thus, the number of arrangements where Eric is to the right of Ortega is equal to the number of arrangements in which Eric is to Ortega's left. Since the total number of arrangements is n, the total number of ways in which Eric can sit to Ortega's right is n/2.

Answer: B