SOLUTION: There are 5 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 thisâ€

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Question 1159556: There are 5 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 by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
(5-1) = 4 performers can be scheduled in 4! = 4*3*2*1 = 24 ways.


The number of all different permutations of 4 objects/items/performers is 4! = 24.

Solved, answered, explained and completed.

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On Permutations,  see introductory lessons
    - Introduction to Permutations
    - PROOF of the formula on the number of Permutations
    - Simple and simplest problems on permutations
    - Special type permutations problems
    - Problems on Permutations with restrictions

    - OVERVIEW of lessons on Permutations and Combinations
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic  "Combinatorics: Combinations and permutations".


Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.