Question 1108155
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<pre>
1-st person can go out at any of 10 bus stops - 10 opportunities.
2-nd person can go out at any of 10 bus stops - 10 independent opportunities.
3-rd person can go out at any of 10 bus stops - 10 independent opportunities.
4-th person can go out at any of 10 bus stops - 10 independent opportunities.
5-th person can go out at any of 10 bus stops - 10 independent opportunities.
6-th person can go out at any of 10 bus stops - 10 independent opportunities.


In all, there are  {{{10^6}}} different ways.


Same number of ways as how many 6-letter words do exist comprising of 10 given letters (symbols) of the alphabet, if letters repetition is allowed.
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I just solved it at least  two times in this forum.


See the lesson 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF =https://www.algebra.com/algebra/homework/Permutations/Combinatoric-problems-for-entities-other-than-permutations-and-combinations.lesson>Combinatoric problems for entities other than 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 lesson is 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.