1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Counting problem.

  1. Feb 27, 2004 #1
    I think i got this answer.

    How many ways are there to distribute five distinguishable objects into three indistinguishable boxes?

    Wouldn't the answer just be [tex]C(5,3)[/tex] because the boxes are indistinguishable? Or do i treat this question the same as if the boxes were distinguishable?
     
  2. jcsd
  3. Feb 27, 2004 #2

    selfAdjoint

    User Avatar
    Staff Emeritus
    Gold Member
    Dearly Missed

    The binomial coefficient [tex]C_3^5[/tex] is right. You could do it as if the boxes were distinguishable and then divide by the number of sequences of them: 3! = 6. and that's what you would get.
     
  4. Feb 27, 2004 #3

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Multiple objects can fit in a box, can't they?
     
  5. Feb 27, 2004 #4
    Unless they are very, very big.
     
  6. Feb 27, 2004 #5
    Yes, multiple objects can fit into a box.

    I think my answer is correct, isn't it?
     
  7. Feb 27, 2004 #6

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    I don't think it is. I think it's quite a way out. The answer is the same as the number of ways to partition 5 into three (possibly empty?) subsets, which is

    [tex]\sum\binom{n}{p}\binom{n}{q}\binom{n}{r}[/tex]

    for all p+q+r = 5 (and possibly with the requirement that p,q,r are strictyl positive.

    To show that your answer is wrong, I think your reasoning that the number of ways to put 5 distinct objects into 1 box is 5 choose 1, ie 5, when obviously it is 1. Also there are many ways of putting 5 balls into 100 boxes, and 5 choose 100 is define to be 0.
     
  8. Feb 27, 2004 #7

    NateTG

    User Avatar
    Science Advisor
    Homework Helper

    There are 5 possible groupings:
    0 0 5 - 1
    0 1 4 - 5
    0 2 3 - 10
    1 1 3 - 10
    1 2 2 - 20

    For a total of 46 possibilities. I don't see a more general formula for this.
     
  9. Feb 27, 2004 #8

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    The answer is the partition number of (5,3) for want of a better term which is what you've worked out, NateG, and does not follow the formula I wrote just there. Anyway, it isn't 5 choose 3. Is
    [tex]\sum\binom{n}{p}\binom{n-p}{q}\binom{n-p-q}{r}[/tex]

    getting better? I think so.

    matt
     
    Last edited: Feb 27, 2004
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Counting problem.
  1. Counting, kinda (Replies: 29)

Loading...