Question

There are eight performers who will present their comedy acts this weekend at a comedy club. One of the performers insists on being the last standup comic of the evening. If this performance request is granted, how many different ways are there to schedule the appearances?

Answer 1

When one performer insists on being last in a lineup of eight, the remaining seven performers can be scheduled in 5,040 different ways, as calculated by the factorial of 7 (7!).

Step-by-step explanation:

If one of the performers insists on being the last standup comic of the evening, we are left with seven other performers whose acts can be scheduled in any order before the last performer goes on stage. Therefore, we need to calculate the number of different permutations for these seven performers.

Since the order of these seven performances matters, we use the formula for permutations without repetition, which is n! (factorial of n), where n is the number of items to arrange. In this case, n is 7, because we have seven performers whose order we need to determine.

The calculation is as follows:

7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5,040

Therefore, there are 5,040 different ways to schedule the appearances of the seven performers before the performer who wishes to go last.