Results 1 to 8 of 8

Math Help - A box contains 36 identical sweets

  1. #1
    Newbie
    Joined
    Nov 2010
    Posts
    10

    A box contains 36 identical sweets

    How many different ways can the sweets be divided among 6 kids so that each of them receives atleast 3 sweets?

    Firstly I'd give each of them 3 sweets, a total of 18.

    Then distribute the other 18 any way possible? Not sure how to go about that though?

    Thanks in advance for any help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,417
    Thanks
    718
    You need to count the number of ways to represent 18 as a sum of 6 natural numbers (possibly equal to zero). See this Wiki page.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,386
    Thanks
    1476
    Awards
    1
    The number of ways the put N identical objects into K different cells is:
    \dbinom{N+K-1}{N}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2010
    Posts
    10
    Quote Originally Posted by Plato View Post
    The number of ways the put N identical objects into K different cells is:
    \dbinom{N+K-1}{N}.
    Ok I did something similar. Here's what I got:

    If you had to find the number of different ways that 36 identical sweets can be divided among 6 kids it would 41Choose5 so the answer is 749,398.

    Ofcourse I want each of the kids to have atleast 3 sweets. So I'd say the answer to my original question is 23Choose5 = 33,649.

    Is that right?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,386
    Thanks
    1476
    Awards
    1
    \dbinom{18+6-1}{18}
    You have already used 18, three to each child.
    You have 18 left. They are identical.
    There are still 6 different children.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Nov 2010
    Posts
    10
    Quote Originally Posted by Plato View Post
    \dbinom{18+6-1}{18}
    You have already used 18, three to each child.
    You have 18 left. They are identical.
    There are still 6 different children.
    and that is the same as 23choose5 which is 33,649.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, Markhor!

    How many different ways can 36 sweets be divided among 6 kids
    so that each of them receives atl east 3 sweets?

    Firstly I'd give each of them 3 sweets, a total of 18.

    Then distribute the other 18 any way possible? . Yes!

    We will distribute the other 18 sweets among the 6 kids.

    For each sweet, we have 6 choices of kids to give it to.

    So there are: . 6^{18} possible distributions.

    . .
    Edit: This is wrong! . . . *blush*

    There are: . 19^5 distributions.
    Last edited by Soroban; November 25th 2010 at 04:51 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,417
    Thanks
    718
    Quote Originally Posted by Soroban View Post
    So there are: . 6^{18} possible distributions.
    It is reasonable to assume that children are numbered but sweets are not. In fact, the OP says that the sweets are identical.

    Now, 6^{18} is the set of all 18-tuples (x_1, ..., x_{18}) where 1\le x_i\le 6 for each i=1,\dots,18. Each x_i says to which kid the ith sweet goes. However, since sweets are identical, we don't care whether sweets #1 through 9 go to kid #1 and those #10 through 18 go to kid #2 or vice versa. So, in our model we count not the number of ordered tuples, but the number of multisets \{x_1,\dots,x_{18}\}. This is \left(\!\tbinom{6}{18}\!\right)=\tbinom{23}{18}=\t  binom{23}{5}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Arithmetic Progression: Sweets
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: May 9th 2011, 04:58 AM
  2. Another ''36 sweets'' counting problem ???
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: November 29th 2010, 10:59 AM
  3. Replies: 3
    Last Post: October 15th 2010, 01:30 AM
  4. When two waves are identical?
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 18th 2009, 02:06 AM
  5. If two identical masses
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: August 16th 2006, 11:55 AM

Search Tags


/mathhelpforum @mathhelpforum