Choose the A students C(16,2)=120 ways.

Choose the B students C(16-2=14,6)=C(14,6)=3003 ways.

Choose the C students C(14-6=8,3)=C(8,3)=56 ways.

Choose the D students C(8-3=5,2)=C(5,2)=10 ways.

Choose the F students C(5-2=3,3)=C(3,3)=1 way.



Answer = (120)(3003)(56)(10)(1) = 201801600



Edwin
 

Answer by ikleyn(52805)   (Show Source): You can put this solution on YOUR website!
 .





This problem is a TWIN to the problem solved 5 minutes ago





https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1204022.html





The same logic works.





     = 120 * 3003 * 56 * 10 = 201801600 different ways.    ANSWER




Solved.





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





To see many similar  (and different)  problems,  solved and explained,  look into the lesson



    - Fundamental counting principle problems 



in this site.





Learn the subject from there.







 

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!
 


In solving a problem like this, it is useful to know that the different grades can be assigned in any order; the result will be the same.  For example, if the grades are assigned from lowest (F) to highest (A), then the calculation is





Of course the last factor  is not needed -- but showing it makes the complete process more clear.


And a third way to find the answer to the question is to count the number of different ways of arranging the 16 grades in a string:


AABBBBBBCCCDDFFF


By a well-known counting principle, the number of distinct arrangements of those letters is





 

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