Question 1043953
.
Hello,I have tried approaching this question but don't know where to begin...
A school chess team consists of 10 students. 
In how many ways can a team of 6 students be formed for three consecutive games, if there should be a different line-up for each game?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
1.  As the first step, let us find how many 6-students teams can be formed of 10 students.

    The answer is: {{{C[10]^6}}} = {{{(10*9*8*7*6*5)/(1*2*3*4*5*6)}}} = 210.


    At this moment I don't know if I should divide by the denominator.
       If we must form a team each time, then I have to divide, and this formula is correct.

       If we must to form a line each time (saying who is playing at the 1-st desk, who is playing at the 2-nd desc and so on) 
             - then I do not have to divide.

       The condition does not allow me to choose between these options - at least as I understand the condition.


       // So the condition is not ideal, by combining these terms "the team" and "line-up" in arbitrary way.  
          The ideal condition should point which of two versions works/is valid.


2. As the second, and final step, let us find in how many ways 3 different 6-student teams can be selected of the total 
   number of 210 different/distinguishable teams.

   The answer is: in 210*209*208 = 9129120 ways.
</pre>

The second opinion is desired.


But the major idea is clear: first find the number of teams/lines.
Then select three of the full number/amount.