Results 1 to 4 of 4

Math Help - Combinatorial analysis

  1. #1
    Junior Member
    Joined
    Mar 2010
    Posts
    60

    Combinatorial analysis

    A set of playing cards consists of 36 cards. These 36 cars are subdivided into 9 types (A,B,C, (...)I) with four cards each. The cards within these subtypes do not differ (A1=A2=A3=A4). How many possibilites are there, to take up 9 cards? The order of taking up cards and the order of the taken up cards do not matter (AAABBCCCD=ABABACDCC).

    I somehow do not get a correct result. First i thought of how man possibilites there are to take 9 cards from 36, but i dont know how to get rid of all possibilites. I tried 36!/27!4!^9, but this is totally wrong...

    thanks for your help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    Quote Originally Posted by Schdero View Post
    A set of playing cards consists of 36 cards. These 36 cars are subdivided into 9 types (A,B,C, (...)I) with four cards each. The cards within these subtypes do not differ (A1=A2=A3=A4). How many possibilites are there, to take up 9 cards? The order of taking up cards and the order of the taken up cards do not matter (AAABBCCCD=ABABACDCC).

    I somehow do not get a correct result. First i thought of how man possibilites there are to take 9 cards from 36, but i dont know how to get rid of all possibilites. I tried 36!/27!4!^9, but this is totally wrong...

    thanks for your help
    Here is a feasible but somewhat laborious way of doing it. Consider, for example, the case where you are given 4 cards from one of the 9 subtypes, 3 from another, and 1 from yet two other subtypes. How many cases of that general sort, "4+3+1+1" are there?

    Well, you can choose the subtype that contributes 4 cards in 9 ways. Having choosen that subtype, you can choose the subtype that contributes 3 cards in 8 ways, and the two subtypes that contribute 1 card each can be chosen in \binom{7}{2} ways. In all, there are 9\cdot 8\cdot\binom{7}{2}=1512 different cases of being dealt 9 cards that follow this general "4+3+1+1" pattern.
    So, I think, it comes down to listing all these possible ways of partitioning the number 9 into a sum of at most 9 non-negative integers ( \leq 4) and, for each of these partitionings, determining how many ways there are to get them by combining the 9 subtypes of cards accordingly.
    Such a list of partitionings might start like this (I am leaving out trailing 0s)

    \begin{array}{lcl}9 &=&4+4+1\\<br />
&=& 4+3+2\\<br />
&=& 4+3+1+1\\<br />
&=& 4+2+2+1\\<br />
&\vdots& \vdots<br />
\end{array}
    To keept this list reasonably small, you should only list the sums that have non-increasing terms (from left to right).
    Last edited by Failure; May 15th 2010 at 04:37 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1
    Quote Originally Posted by Schdero View Post
    A set of playing cards consists of 36 cards. These 36 cars are subdivided into 9 types (A,B,C, (...)I) with four cards each. The cards within these subtypes do not differ (A1=A2=A3=A4). How many possibilites are there, to take up 9 cards?
    If we expand (1+x+x^2+x^3+x^4)^9 the coefficient of x^9 is 19855.
    That is the answer to your question done with a generating function.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    Quote Originally Posted by Plato View Post
    If we expand (1+x+x^2+x^3+x^4)^9 the coefficient of x^9 is 19855.
    That is the answer to your question done with a generating function.
    That's a great way of doing it, except for two possible problems: (1) that the OP does not know about the use of generating functions in enumerative combinatorics, and, (2) that he does not have access to a CAS to determine the coefficient of x^9 so very easily as the above statement seems to suggest.

    In fact, it seems to me that if he does not have access to a CAS (or is not allowed to use a CAS) for that purpose, he might as well use the laborious procedure that I have suggested.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Combinatorial Probability
    Posted in the Statistics Forum
    Replies: 7
    Last Post: December 7th 2011, 03:32 PM
  2. Probability - Combinatorial
    Posted in the Statistics Forum
    Replies: 5
    Last Post: August 15th 2009, 07:26 AM
  3. combinatorial sum
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: May 12th 2009, 10:46 AM
  4. Combinatorial Analysis
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 5th 2009, 10:18 AM
  5. Combinatorial Analysis
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 23rd 2009, 03:11 PM

Search Tags


/mathhelpforum @mathhelpforum