Question 1136227
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            I am not familiar with this train root and don't know exactly how many stops 
            are there from Cambridge to London.


            And the problem do not point it exactly.  So I will assume 
            that there are 9 stations, counting from the next from Cambridge to London INCLUSIVELY.


            Also,  I will assume that all 6 persons entered at Cambridge,  but are free to leave the train at any other station.


            If it is not so, then my solution needs to be corrected.  (Sorry)



<pre>
Of 6 persons entering at Cambridge, each is free to have a ticket to any of 9 stations.


It gives  9 opportunities in each of 6 positions;  so, there are  {{{9^6}}}  different sets of tickets possible.    <U>ANSWER</U>
</pre>

<pre>
I can reformulate this problem in other form:

    Let  9 letters A, B, C, D, E, F, G, H, I represent the names of the stations starting from next after Cambridge to London inclusively.


    Then every of 6 person may have in his hand any of these 9 letters as his ticket.


    Then the problem's question is: how many 6-letter words do exist written in this 9-letter alphabet.


    The  <U>ANSWER</U>  is  {{{9^6}}}  words.

    The same answer as in the previous solution.
</pre>

Surely, &nbsp;both solutions are the same, &nbsp;simply expressed in different terms.


Solved.



My opinion is that this problem &nbsp;(to be in style of Conan Doyle's  Sherlock Holmes and Dr. Watson) &nbsp;could be 

(and should be) &nbsp;formulated in more precise terms . . . 


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To see other similar problem solved, &nbsp;look into 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>, &nbsp;&nbsp;Problem 7 

in this site.