SOLUTION: There are 7 performers who will present their comedy act this weekend at a comedy club. One of the performers insists on being the last stand-up comic of the evening. If this perfo

Algebra ->  Permutations -> SOLUTION: There are 7 performers who will present their comedy act this weekend at a comedy club. One of the performers insists on being the last stand-up comic of the evening. If this perfo      Log On


   

Question 1110531: There are 7 performers who will present their comedy act 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(52779) About Me  (Show Source):
You can put this solution on YOUR website!
.
6! = 6*5*4*3*2*1 = 720 ways.

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On Combinations and Permutations see the lessons
    - Introduction to Permutations
    - PROOF of the formula on the number of Permutations
    - Problems on Permutations
    - 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.