Question 1185993
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<pre>

There are 14! permutations (listings) of 14 items.


14! = 1*2*3*4*5*6*7*8*9*10*11*12*13*14 = 87178291200.


<U>ANSWER</U>.  87178291200  different listings are possible.



Any of 14 teams can be listed first.

Any of remaining 13 teams can be listed second.


and so on . . . 
</pre>

Solved and explained.


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This problem is on PERMUTATIONS.



On Permutations, &nbsp;see introductory lessons 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =http://www.algebra.com/algebra/homework/Permutations/Introduction-to-Permutations.lesson>Introduction to Permutations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =http://www.algebra.com/algebra/homework/Permutations/PROOF-of-the-formula-on-the-number-of-permutations.lesson>PROOF of the formula on the number of Permutations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =http://www.algebra.com/algebra/homework/Permutations/Problems-on-Permutations.lesson>Simple and simplest problems on permutations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =https://www.algebra.com/algebra/homework/Permutations/OVERVIEW-the-lessons-on-Permutations-and-Combinations.lesson>OVERVIEW of lessons on Permutations and Combinations</A>

in this site.


Also, &nbsp;you have this free of charge online textbook in ALGEBRA-II in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this online textbook under the topic &nbsp;"<U>Combinatorics: Combinations and permutations</U>". 



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.