Results 1 to 8 of 8

Math Help - Combinations

  1. #1
    Member
    Joined
    May 2006
    Posts
    148
    Thanks
    1

    Combinations

    This is a question from an assignment given in a Computational Algorithm class.

    Ten research labs are attempting to determine the efficacy of proposed drugs in treating cancer. All the drugs are ordered from a pharmaceutical company (all ordered from the same one). Each of the labs want to order 1 drug to test. The choices are the following: Xenr, Phum, and Tsela.

    The questions:

    a.) How many orders are possible?

    b.) Xenr appears promising. How many orders are there where at least 3 labs order Xenr?

    c.) The pharmaceutical company announces it's running low on Tsela and only has enough for 2 labs. How many orders from the 10 labs can the company fill?

    I am not sure which combinatoric principals would be used. a.) apppears it'd just be 10^3.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by fifthrapiers View Post
    This is a question from an assignment given in a Computational Algorithm class.

    Ten research labs are attempting to determine the efficacy of proposed drugs in treating cancer. All the drugs are ordered from a pharmaceutical company (all ordered from the same one). Each of the labs want to order 1 drug to test. The choices are the following: Xenr, Phum, and Tsela.

    The questions:

    a.) How many orders are possible?
    as you said, the answer is 3^10

    b.) Xenr appears promising. How many orders are there where at least 3 labs order Xenr?
    three labs order Xenr, there are seven more labs that can order 3^7 different combinations. Therefore there are 3^7 combinations

    c.) The pharmaceutical company announces it's running low on Tsela and only has enough for 2 labs. How many orders from the 10 labs can the company fill?
    Find the number of possible combinations where 3 or more labs by Tsela, which is 3^7 (which equals 2187)

    Now subtract that from the total possible combinations: 59049-2187=56862

    so there are 56862 possible combinations if no more than 2 labs can buy Tsela.

    make note: 3^10=59049
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2006
    Posts
    148
    Thanks
    1
    Thanks for the reply; I thought my answer (and subsequently yours), appeared too simple! Apparently this is not the correct answer, according to the professor. She said she could understand how that answer was created, and said to add,

    "All of the 10 labs are @ the same univ. and the univ. places the "drugs" request @ the pharmac. company."

    According to the professor, I should be using the r-selection method for this; that is, order is not important, and repetition is allowed. I believe it'll use the equation C(r+n-1, r)..
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,705
    Thanks
    1637
    Awards
    1
    Quote Originally Posted by fifthrapiers View Post
    Ten research labs are attempting to determine the efficacy of proposed drugs in treating cancer. All the drugs are ordered from a pharmaceutical company (all ordered from the same one). Each of the labs want to order 1 drug to test. The choices are the following: Xenr, Phum, and Tsela.
    The questions: a.) How many orders are possible?
    b.) Xenr appears promising. How many orders are there where at least 3 labs order Xenr?
    c.) The pharmaceutical company announces it's running low on Tsela and only has enough for 2 labs. How many orders from the 10 labs can the company fil.
    I can tell you that this is a very poorly worded problem!
    I say that as someone who has been of many editorial boards, regional and national, for contest questions.

    In this case, I would have suggested this revision: “The university’s buying agent has to order drugs from one pharmaceutical company. The agent is buying one of enr, Phum, and Tsela for each of ten labs on campus.
    a.) How many orders are possible?
    b.) Xenr appears promising. How many orders are there where at least 3 labs order Xenr?
    c.) The pharmaceutical company announces it's running low on Tsela and only has enough for 2 labs. How many orders from the 10 labs can the company fill?
    Now your instructor’s suggestion makes perfect sense!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2006
    Posts
    148
    Thanks
    1
    Indeed, well not to me any way .

    Tried working on c.): is it (10 choose 9) * (9 choose 3) ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,705
    Thanks
    1637
    Awards
    1
    c.) The Pharmaceutical Company announces it's running low on Tsela and only has enough for 2 labs. How many orders from the 10 labs can the company fill?

    If we take that to mean “How many orders from the 10 labs can the company fill with no more than 2 Tselas in the order” then that is the complement of at least three Tselas in the order. [C(N,k) means N choosing k.] The number of total possible orders is C(10+3-1,10)=C(12,10). The number of total possible orders with at least three Tselas is C(7+3-1,7), think of going ahead putting three in the ‘Tsela box’ and distributing the remaining seven in all three boxes. Thus, C(12,10)-C(9,7) is the total number of orders with no more than 2 Tselas.
    Last edited by Plato; September 14th 2006 at 12:21 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    May 2006
    Posts
    148
    Thanks
    1
    Okay, I thought I had a and b right thinking c was too, but now it just confused me with your answer. Initially, I thought a.) would just be 3^10. I mean, there are 10 ways to choose the 1st drug, then 10 ways for the 2nd, and then 10 ways for the 3rd.

    And for b, with 7 remaining labs Quick's answer of 3^7 seemed logical, but there has to be some combinatorics, so something's wrong.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,705
    Thanks
    1637
    Awards
    1
    It is fairly clear from what your instructor said that she is expects you to use the multi-selection rule: C(N+k-1,k) ways to make N choices from k different items. You are ordering ten from three different drugs. The labs are not a consideration here. We are simply interested in how many different orders are possible of ten from three.
    Last edited by Plato; September 15th 2006 at 04:42 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Combinations in a set
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: September 9th 2010, 06:19 AM
  2. How many combinations are possible?
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: July 23rd 2009, 07:53 PM
  3. Combinations
    Posted in the Statistics Forum
    Replies: 2
    Last Post: May 5th 2008, 08:28 AM
  4. How many combinations..?
    Posted in the Algebra Forum
    Replies: 9
    Last Post: May 2nd 2008, 10:34 AM
  5. combinations
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 27th 2008, 09:00 AM

Search Tags


/mathhelpforum @mathhelpforum