Results 1 to 3 of 3

Math Help - Choosing dishes and sharing...

  1. #1
    Newbie
    Joined
    Nov 2012
    From
    Arizona
    Posts
    8

    Choosing dishes and sharing...

    Hi,

    Here is a problem I'm having a hard time wrapping my head around.

    10 people go to a restaurant to order 10 dishes out of 50 distint dishes. If people can share the dishes among themselves then how many distinct eating experiences are possible?

    First of, there are 2 cases for whether repetition is allowed of not. In the first case, if repetition is not allowed then there are 50P10 distinct experiences, else it is 5010. (Permutation because the order of dishes matters in creating distinct experiences.)

    How to consider sharing of the 10 dishes among the 10 people? In the no repetition case, 10 distinct dishes can be shared among 10 people in 1010 ways, so the total is 50P10 X 1010 ? What about the case where when repetition allowed, i.e., all people can have the same dish?

    Any help is appreciated.
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,160
    Thanks
    348

    Re: Choosing dishes and sharing...

    The fact that they can share dishes implies that order is not important - if Andy chooses dish A and Bill dish B it's the same experience as Andy choosing B and Bill choosing A, because they could exchange dishes if they want (the ultimate in sharing). Similarly if Andy and Bill both choose A and Charlie chooses B, I think that's the same experience as Andy choosing A and Bill and Charlie both choosing B. So I interpret this question to mean: how many ways can 10 out of 50 dishes be chosen? If there are no repeats then that would be 50C10 - note that we use combination, not permutation, because the order of the dishes is not important. But supoose there is a repeat, so that only 9 distict dishes are selecetd by the ten people - the numbner of ways that 9 dishes can be picked is 50C9. And so on for cases where only 8 distinct dishes are picked, or 7, etc. So in total you have

    50C10 + 50C9 + 50C8 + ...+ 50C1 distinct combinations.
    Last edited by ebaines; February 12th 2013 at 02:17 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2012
    From
    Arizona
    Posts
    8

    Re: Choosing dishes and sharing...

    Thanks for the quick response. Your interpretation makes sense (Sharing is equivalent to choosing since order doesn't matter anymore). I was thinking that sharing should mean more distinct experiences.

    So in summary, (please correct me if I am wrong)

    No repetition: 50P10 if no sharing and 50C10 if there is sharing.
    With repetition: 50^10 if no sharing and (50C10 + 50C9 + 50C8 + ...+ 50C1) if there is sharing

    Thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Math Knowledge Sharing
    Posted in the New Users Forum
    Replies: 0
    Last Post: January 21st 2013, 06:07 PM
  2. Secret Sharing
    Posted in the Math Challenge Problems Forum
    Replies: 0
    Last Post: November 18th 2012, 01:22 AM
  3. help sharing settings accurately.
    Posted in the Algebra Forum
    Replies: 1
    Last Post: June 29th 2009, 03:51 AM
  4. Sharing a Math Forum
    Posted in the Pre-Calculus Forum
    Replies: 0
    Last Post: November 28th 2008, 09:58 AM
  5. 5 pirates sharing the booty
    Posted in the Math Challenge Problems Forum
    Replies: 25
    Last Post: September 12th 2007, 09:00 PM

Search Tags


/mathhelpforum @mathhelpforum