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pythagoras123 Group Title

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.

  • 2 years ago
  • 2 years ago

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  1. FoolForMath Group Title
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    250 is the answer?

    • 2 years ago
  2. pythagoras123 Group Title
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    I'm not too sure of the answer, but 250 might be too big...

    • 2 years ago
  3. m_charron2 Group Title
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    Do we care about the order of the presents in the boxes or it's simply the number of presents per box that matter?

    • 2 years ago
  4. FoolForMath Group Title
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    In distributing the boxes, is it possible to give 2 boxes to one student?

    • 2 years ago
  5. pythagoras123 Group Title
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    The order does matter. Remember that we are giving to 3 different students.

    • 2 years ago
  6. FoolForMath Group Title
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    If YES then the answer is 250 if No then the answer should be 25.

    • 2 years ago
  7. pythagoras123 Group Title
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    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.

    • 2 years ago
  8. pythagoras123 Group Title
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    By working, I mean the steps taken to get the answer.

    • 2 years ago
  9. FoolForMath Group Title
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    Sure, \( S(5,3) \times \binom{3-1}{3-1} = 25 \) Where S() is Stirling number of second kind.

    • 2 years ago
  10. Ishaan94 Group Title
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    Can't we do this in conventional way? hehe... No Stirling Number.

    • 2 years ago
  11. m_charron2 Group Title
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    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...

    • 2 years ago
  12. m_charron2 Group Title
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    oh, nvm that, I'm possibly counting some combinations twice (if not more)

    • 2 years ago
  13. Ishaan94 Group Title
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    ohh wait...

    • 2 years ago
  14. Ishaan94 Group Title
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    ohh hehehe (5P3*3P2)

    • 2 years ago
  15. Ishaan94 Group Title
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    5*4*3*3*2, hmm but m_charron2 got 3^2 there. WHY?

    • 2 years ago
  16. m_charron2 Group Title
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    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

    • 2 years ago
  17. Ishaan94 Group Title
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    so, instead of 3^2, 3P2? but still (5*4*3)*6 > 25. How did foolformath get 25?

    • 2 years ago
  18. Ishaan94 Group Title
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    @Zarkon

    • 2 years ago
  19. m_charron2 Group Title
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    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

    • 2 years ago
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