Results 1 to 10 of 10

Math Help - Counting

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    69

    Question Counting

    A candy factory has an endless supply of red, orange, yellow, green, blue, and violet jelly beans. The factory packages the jelly beans into jars of 84 jelly beans each. As a marketing gimmick, the factory guarantees that no two jars have exactly the same number of red, orange, yellow, green, blue, and violet jelly beans. What is the maximum number of jars the factory can produce?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,657
    Thanks
    1609
    Awards
    1
    Quote Originally Posted by essedra View Post
    A candy factory has an endless supply of red, orange, yellow, green, blue, and violet jelly beans. The factory packages the jelly beans into jars of 84 jelly beans each. As a marketing gimmick, the factory guarantees that no two jars have exactly the same number of red, orange, yellow, green, blue, and violet jelly beans. What is the maximum number of jars the factory can produce?
    How many non-negative integral solutions are there to this equation are there:
    r+o+y+g+b+v=84~?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2009
    Posts
    69
    Quote Originally Posted by Plato View Post
    How many non-negative integral solutions are there to this equation are there:
    r+o+y+g+b+v=84~?
    I didn't get what you mean...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2009
    Posts
    69
    I tried to solve it like this:

    6x6x6x6x6x6x6x6x6x6x6x6x6x6x6........ (84 times) =

    6^84 = 2 + 65 zeros

    which is same as

    2x10^65

    but the answer was incorrect...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,657
    Thanks
    1609
    Awards
    1
    Quote Originally Posted by essedra View Post
    I didn't get what you mean...
    Somewhere in you text-material there must be a section of making multi-selections.
    How many ways to put N identical objects into K different cells: \binom{N+K-1}{N}.

    So when I asked about many solutions to r+o+y+g+b+v=84,
    that is equivalent putting 84 identical ones into six different colors.
    Each solution is one way the candy factory can package the 84 jelly beans.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Mar 2009
    Posts
    69
    I still couldn't solve the problem, because i'm confused about the "k" you're talking about...are order and repetition important?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Mar 2009
    Posts
    69
    I would be really grateful if you could show me the way to solve this problem...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,657
    Thanks
    1609
    Awards
    1
    Quote Originally Posted by essedra View Post
    I still couldn't solve the problem, because i'm confused about the "k" you're talking about...are order and repetition important?
    Let me give you an example.
    Suppose that an ice cream shop has 21 flavors.
    It runs a special offer of a plate of 4 scoops of any combinations.
    How many different offerings are there?
    Answer: \binom{4+21-1}{4}=\frac{24!}{(4!)(20!)}

    That is the number of ways to put 4 identical objects (scoops) into 21 (flavors) different cells.
    Note that the 4 can be all different, all the same, or say two of one and two different.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Mar 2009
    Posts
    69
    so I applied what you said...

    89!/(6!*84! ) = 6917940,3333333333333333333333333

    but the result was incorrect...
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,657
    Thanks
    1609
    Awards
    1
    Quote Originally Posted by essedra View Post
    so I applied what you said...

    89!/(6!*84! ) = 6917940,3333333333333333333333333

    but the result was incorrect...
    I don't think that you have caculated that correctly.
    \binom{84+6-1}{84}=\binom{89}{84}=41507642
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. counting 1
    Posted in the Statistics Forum
    Replies: 1
    Last Post: September 8th 2011, 08:01 AM
  2. Counting
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: May 9th 2010, 04:34 AM
  3. Counting
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: March 9th 2010, 05:19 PM
  4. help on Counting
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: October 25th 2009, 08:51 AM
  5. Counting
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 10th 2009, 05:32 PM

Search Tags


/mathhelpforum @mathhelpforum