SOLUTION: (A)lex, (B)rian, (C)laire, and (D)ana, have 4 tickets to the play. Use the fundamental counting principal and the slot (blank) method to answer the question In how many ways can

Question 1169540: (A)lex, (B)rian, (C)laire, and (D)ana, have 4 tickets to the play. Use the fundamental counting principal and the slot (blank) method to answer the question
In how many ways can they be arranged so that Alex and Brian are not sitting next to each other?
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

            The instruction in your post is IRRELEVANT and INCORRECT.

            It only CONFUSE the reader and is a REAL OBSTACLE to the solution.

            It makes VERY BAD IMPRESSION about the person who composed this problem,
            because it shows his (or her) EXTREMELY LOW understanding/knowledge of the subject and the problem.

Solution

The idea is to subtract from the total number of all possible permutations of 4 objects 

the number of permutations where persons A and B are together.


The total number of permutations of 4 objects is 4! = 4*3*2*1 = 24.


To determine the number of permutations of 4 persons where A and B are together, we consider A and B as one object,
but we shoud account that A and B can sit in two different configurations (AB) and (BA).


Therefore, the answer to the problem's question is  


    4! - 2*3! = 24 - 2*6 = 24 - 12 = 12.

Solved.

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