Question 1127705
.
<pre>
Any of 9 players can stand first.                (9 options)

Any of remaining 8 players can stand second.     (8 options)

Any of remaining 7 players can stand third       (7 options)

    . . .    and so on   . . . 


In all, there are  9*8*7*6*5*4*3*2*1  ways = 9! ways = 362880 ways to form different lineup.
</pre>

----------------


It is the number of all possible permutations of 9 objects.


On Permutations, &nbsp;see the 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>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.