Question

There are 9

performers who will present their comedy acts this weekend at a comedy club. One of the performers insists on being the last​ stand-up comic of the evening. If this​ performer's request is​ granted, how many different ways are there to schedule the​ appearances?

Answer 1

This can be solved by finding the number of permutations of the eight performers who will all perform before the one who insists on being the last.
$8P8=8\times7\times6\times5\times4\times3\times2\times1=40,320\ ways$

Answer 2

Answer:

Total number of different ways to schedule the​ appearances is 40320.

Step-by-step explanation:

Total number of performers = 9.

One of the performers insists on being the last​ stand-up comic of the evening. We have to arrange one performer in last.

$^1P_1=\frac{1!}{(1-1)!}=\frac{1!}{1}=1$

We have to arrange the remaining performers. The number of remaining performer is

$9-1=8$

Total number of ways to arrange 8 members is

$^8P_8=\frac{8!}{(8-8)!}=\frac{8!}{1}=8\times 7\times 6\times 5\times 4\times 3\times 2\times 1=40320$

Total number of ways is

$40320\times 1=40320$

Therefore, total number of different ways to schedule the​ appearances is 40320.