Question 1193209
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<pre>

From the context, all 6 people are equally qualified to take any of the 6 position.


So, any of the 6 people for the first position;

any of remaining 5 people for the second position;

any of remaining 4 people for the third position,


. . . and so on . . . 


giving  6*5*4*3*2*1  6! = 720 different ways.    <U>ANSWER</U>
</pre>

Solved.


There are 6! = 720 possible permutations.


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This problem is on &nbsp;PERMUTATIONS.


On &nbsp;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>

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.