Let’s solve a combination problem. There are eight speakers in a function whose turns come at random. What is the probability for the first three, A, B, and C, to speak such that A speaks before B and B before C?
Focus on A, B and C first and arrange the rest around them. The number of ways to arrange A, B, and C in that order among eight is 8C3. The number of ways of arranging the other five is 5! This means that the number of ways to set the three and five is 8C3 x 5! That forms the numerator. The denominator becomes the total number of ways of arranging eight people to speak, which is 8!
Therefore, the required probability becomes 8C3 x 5! / 8! = (8! x 5!) /(3! x 5! x 8!) = 1/3! = 1/6.
