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Ethan
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gmatacer
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deowl
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Ethan
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Question 1:

Let A,B,C be students and C1, C2 be class rooms

C1 C2
0 ABC
A BC
B AC
C AB
AB C
BC A
CA B
ABC 0

Total possible ways = 8. However, can someone please explain this in a better way rather than listing all the posibilities. As this approach for Question 2 becomes quite cumbersome. Also, please explain in detail...
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Ethan
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Deowl,
Have we taken into consideration that there is a possiblity of no student being assigned to a class room. What I am not sure at this point is, does n^m take care of possibility of students not assigned to a classroom?

Thanks in advance.
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deowl
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Absolutely. Since each student has equal probability to be assigned to any of the classes, all students could be assigned to one class so other classes remain empty.
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Ethan
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The OA for Question 2 is 36 and I digged this question out from some archives. And I have no clue how it can be 36...Any thoughts....
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deowl
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Ethan

I think you have omitted one important condition from the second question.
( actually this makes sense since who would ask the same question with such a minor modification twice )
The condition is that no rooms should remain empty. In this case we resolve it
in the following way:

1. # of possibilities for room with 2 students: 3
2. # of possibilities to find that lucky couple: 6
3. # of arrangements for remaining students: 2

Total: 3 * 6 * 2 = 36



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