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A school assigns students to small classrooms in such a way [#permalink]
 20 Apr 2006, 05:46
A school assigns students to small classrooms in such a way that some of the classrooms can be empty and more than one student can be assigned to a classroom.
Question 1)In how many ways can the school assign 3 students to 2 different small classrooms?
Question 2)In how many ways can the school assign 4 students to 3 different small classrooms?
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Question 1)In how many ways can the school assign 3 students to 2 different small classrooms?
2[(3C3 * 3C0) + (3C2 + 1)]
2(1+4)
10
Question 2)In how many ways can the school assign 4 students to 3 different small classrooms?
2[(4C2 * 4C2) + (4C1 * 4C3) + (4C4 * 4C0)]
2(36 + 16 + 1)
2(53)
106
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1. 8 ( 2^3 )
2. 81 ( 3^4 )
Each student can be assigned to one of X classrooms, so if there are
Y students they can be assigned in X^Y ways.
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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...
Last edited by Ethan on 20 Apr 2006, 07:03, edited 1 time in total.
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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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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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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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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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close
