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anonymous
 4 years 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.
anonymous
 4 years 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.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm not too sure of the answer, but 250 might be too big...

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The order does matter. Remember that we are giving to 3 different students.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If YES then the answer is 250 if No then the answer should be 25.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0With 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0By working, I mean the steps taken to get the answer.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can't we do this in conventional way? hehe... No Stirling Number.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh, nvm that, I'm possibly counting some combinations twice (if not more)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.05*4*3*3*2, hmm but m_charron2 got 3^2 there. WHY?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The 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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oups, 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
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