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A teacher has 5 different presents to share among 3 students. He distributes the presents in three identical boxes (that cannot be distinguished). There is at least one present in each box. Find the number of ways that the teacher can distribute the presents.
 one year ago
 one year ago
A teacher has 5 different presents to share among 3 students. He distributes the presents in three identical boxes (that cannot be distinguished). There is at least one present in each box. Find the number of ways that the teacher can distribute the presents.
 one year ago
 one year ago

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FoolForMathBest ResponseYou've already chosen the best response.0
250 is the answer?
 one year ago

pythagoras123Best ResponseYou've already chosen the best response.0
I'm not too sure of the answer, but 250 might be too big...
 one year ago

m_charron2Best ResponseYou've already chosen the best response.0
Do we care about the order of the presents in the boxes or it's simply the number of presents per box that matter?
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
In distributing the boxes, is it possible to give 2 boxes to one student?
 one year ago

pythagoras123Best ResponseYou've already chosen the best response.0
The order does matter. Remember that we are giving to 3 different students.
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
If YES then the answer is 250 if No then the answer should be 25.
 one year ago

pythagoras123Best ResponseYou've already chosen the best response.0
With regard to your previous question, the answer is no. By the way, could you show me how you get 25? I just need the working, I'm not so particular about the answer.
 one year ago

pythagoras123Best ResponseYou've already chosen the best response.0
By working, I mean the steps taken to get the answer.
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.0
Sure, \( S(5,3) \times \binom{31}{31} = 25 \) Where S() is Stirling number of second kind.
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
Can't we do this in conventional way? hehe... No Stirling Number.
 one year ago

m_charron2Best ResponseYou've already chosen the best response.0
I must be extremely off, I got 540 different ways :S 5 * 4 * 3 to put one in each box, 3 * 3, because the remaining two can be anywhere in the 3 boxes. 60*9 = 540 Conventionnal way? like buying the right number of presents? Lousy cheapskate teacher...
 one year ago

m_charron2Best ResponseYou've already chosen the best response.0
oh, nvm that, I'm possibly counting some combinations twice (if not more)
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
5*4*3*3*2, hmm but m_charron2 got 3^2 there. WHY?
 one year ago

m_charron2Best ResponseYou've already chosen the best response.0
The reasoning behind my answer was 2part : 1st, all boxes had to have at least one present in it, so 5 P for the first, 4 for the second, 3 for the third. Then, the remaining 2 presents can be in any box, even the 2 being in the same box, so 3^2. But it's wrong, because you get combinations twice, present1 and present2 in box1 and present2 and present1 in box1 are the same thing
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
so, instead of 3^2, 3P2? but still (5*4*3)*6 > 25. How did foolformath get 25?
 one year ago

m_charron2Best ResponseYou've already chosen the best response.0
oups, revised hand counting : 5 possibilities for 1 present being in box1 * 13 combinations for box2 + 10 possibilities for 2 presents being in box in * 6 combinations for box2 + 10 possibilities for 3 presents being in box 1 * 2 combinations for box2 for a total of 145 possibilities
 one year ago
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