Consider the AB combination as a single letter. Then there are 4!=24 ways to arrange the letters AB,C,D,E.
Since A-B can be arranged in two ways, there are 2*4! = 48 ways.
You are given the set of letters ( A, B, C, D, E ). What is the probability that in a random five-letter string (in which each letter appears exactly once, and with all such strings equally likely) the letters A and B are next to each other?
3 answers
but what is the denominator?
Consider the AB combination as a single letter. Then there are 4!=24 ways to arrange the letters AB, C, D, E.
Since A-B can be arranged in two ways, there are 2*4! = 48 ways.
Since the total of combinations is 5!= 120
The answer will be 48/120 = 0.4
Since A-B can be arranged in two ways, there are 2*4! = 48 ways.
Since the total of combinations is 5!= 120
The answer will be 48/120 = 0.4