pythagoras123
  • pythagoras123
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.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
250 is the answer?
pythagoras123
  • pythagoras123
I'm not too sure of the answer, but 250 might be too big...
anonymous
  • anonymous
Do we care about the order of the presents in the boxes or it's simply the number of presents per box that matter?

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anonymous
  • anonymous
In distributing the boxes, is it possible to give 2 boxes to one student?
pythagoras123
  • pythagoras123
The order does matter. Remember that we are giving to 3 different students.
anonymous
  • anonymous
If YES then the answer is 250 if No then the answer should be 25.
pythagoras123
  • pythagoras123
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.
pythagoras123
  • pythagoras123
By working, I mean the steps taken to get the answer.
anonymous
  • anonymous
Sure, \( S(5,3) \times \binom{3-1}{3-1} = 25 \) Where S() is Stirling number of second kind.
anonymous
  • anonymous
Can't we do this in conventional way? hehe... No Stirling Number.
anonymous
  • anonymous
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...
anonymous
  • anonymous
oh, nvm that, I'm possibly counting some combinations twice (if not more)
anonymous
  • anonymous
ohh wait...
anonymous
  • anonymous
ohh hehehe (5P3*3P2)
anonymous
  • anonymous
5*4*3*3*2, hmm but m_charron2 got 3^2 there. WHY?
anonymous
  • anonymous
The reasoning behind my answer was 2-part : 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
  • anonymous
so, instead of 3^2, 3P2? but still (5*4*3)*6 > 25. How did foolformath get 25?
anonymous
  • anonymous
@Zarkon
anonymous
  • anonymous
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

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