Results 1 to 4 of 4

Math Help - divide into groups or arrange into groups

  1. #1
    Member
    Joined
    May 2009
    Posts
    91

    divide into groups or arrange into groups

    Hi folks,

    My maths book tells me that if I want to divide 12 people into 3 groups containing 2,3 and 7 people the following logic applies:

    The two people of the first group are chosen in [HTML]<sup>12</sup>C<sub>2</sub>[/HTML] ways i.e. 12!/10!2!

    The three people of the second group are chosen in [HTML]<sup>10</sup>C<sub>3</sub>[/HTML] ways i.e. 10!/7!3!

    and the third group contains the remainder which can be chosen in only 1 way. So we have:

    12! 10! / 10! 2! 7! 3! = 12! / 2!3!7! = 7920 ways.

    Then it gives a general principle supporting this logic. "The number of ways of dividing (p+q+r) unlike things into 3 groups containing p, q and r things respectively is: (p+q+r)! / p!q!r!

    A couple of pages on my maths book asks: how many ways can 12 people be arranged into groups if there are to be 2 groups of six people?

    I would have used the same technique i.e. 12! / 6! 6! = 924

    But the books provides the following logic:

    The first group of 6 can be selected in [HTML]<sup>12</sup>C<sub>6</sub>[/HTML] ways and the second group of 6 is then automatically made up of those not in the first group, so there is only one way of selecting the second group, but since the [HTML]<sup>12</sup>C<sub>6</sub>[/HTML] ways include the case when a1,a2,a3...a6 are in the first group and a7, a8... a12 are in the second group and vice versa we divide by 2 (i.e. by the number of ways the two groups can be permutated amongst themselves) and the result is

    12! / 6! 6! 2 = 462.

    So my question is this: how do you divide into groups? using method 1 or method 2?
    Last edited by s_ingram; June 16th 2009 at 01:19 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1
    Quote Originally Posted by s_ingram View Post
    A couple of pages on my maths book asks: how many ways can 12 people be arranged into groups if there are to be 2 groups of six people? I would have used the same technique i.e. 12! / 6! 6! = 924

    But the books provides the following logic:
    The first group of 6 can be selected in [HTML]<sup>12</sup>C<sub>6</sub>[/HTML] ways and the second group of 6 is then automatically made up of those not in the first group, so there is only one way of selecting the second group, but since the [HTML]<sup>12</sup>C<sub>6</sub>[/HTML] ways include the case when a1,a2,a3...a6 are in the first group and a7, a8... a12 are in the second group and vice versa we divide by 2 (i.e. by the number of ways the two groups can be permutated amongst themselves) and the result is 12! / 6! 6! 2 = 462.
    So my question is this: how do you divide into groups? using method 1 or method 2?
    In the first example the groups were of different sizes. Therefore, they were labeled in a sense: 2,3,7
    In the second case we simply have two groups of six each. The groups are unlabeled.

    Examples
    Divide a group of 15 people into subgroups of 2, 3, 4, 6; \frac{15!}{(2!)(3!)(4!)(6!)} ways.

    Divide a group of 15 people into three subgroups of five each; \frac{15!}{(5!)^3(3!)} ways.

    Divide a group of 15 people into three subgroups of 3, 6, 6; \frac{15!} {(3!) (6!)^2(2!)} ways.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2009
    Posts
    91

    Red face

    Hi Plato (great name by the way!)

    thanks for your answer. I see what you are saying but I don't understand it.

    If I have twelve people and I am asked to divide them into subgroups of 2, 3 and 7, I can do this in

    12! / 2! 3! 7! ways

    If I am asked to divide 12 people into 3 subgroups of 4, I do this in

    12! / 4! 4! 4! 3! ways

    both the above give the correct result, but the reasoning is the problem! To me there seems no explanation for the extra 3! in the second case. I know it corresponds to the fact that the 3 groups can be "permutated amoung themselves" in 3! ways and that in the first case each group is a single subgroup with no cross terms - but it all seems somehow unconvincing. The fact that in the second case the groups are of the same size shouldn't change the character of the problem. It seems to be about our labelling system rather than the underlying nature of the problem. We are after all just splitting 12 people into 3 groups!
    If you can shed any light on why the fact that a different sized group makes a difference I would be very pleased to hear!

    best regards and thanks for your time
    Simon
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1
    Quote Originally Posted by s_ingram View Post
    It seems to be about our labeling system rather than the underlying nature of the problem.
    Simon, labeling is the underlying essence of this problem.
    Suppose we have twelve students in a prep school: A,B,C,D,E,F,G,H,I,J,K,\&~L.
    We need to divide these in three study groups of four each: one studies chemistry, one studies calculus, and the third studies literature.
    One possible groupings is \underbrace {\left\{ {{\text{A,C,E,L}}} \right\}}_{{\text{chemistry}}},\underbrace {\left\{ {{\text{B,E,F,H}}} \right\}}_{{\text{calculus}}},\underbrace {\left\{ {{\text{D,G,I,J}}} \right\}}_{{\text{literature}}}.

    Here is another: \underbrace {\left\{ {{\text{A,C,E,L}}} \right\}}_{{\text{ calculus }}},\underbrace {\left\{ {{\text{B,E,F,H}}} \right\}}_{{\text{ literature }}},\underbrace {\left\{ {{\text{D,G,I,J}}} \right\}}_{{\text{ chemistry }}}.

    Notice that the groups have the exact same content but different labels.
    Clearly the two are really different assignments.
    I should think it may make a real difference to Mr. A if he studies calculus or literature.

    On the hand, divide them into three groups for a work detail these two are the same:
    \left[ {\left\{ {{\text{A,C,E,L}}} \right\},\left\{ {{\text{B,E,F,H}}} \right\},\left\{ {{\text{D,G,I,J}}} \right\}} \right]\;\& \;\left[ {\left\{ {{\text{B,E,F,H}}} \right\},\left\{ {{\text{D,G,I,J}}} \right\},\left\{ {{\text{A,C,E,L}}} \right\}} \right].
    They are the same grouping but not labeled.
    Again it may make a difference to Mr. B who is in his group but not the ordering of the three groups.
    Thus we remove the labels by dividing by 3!.

    The number of ways to divide a collection of thirty-two different pieces of jewelry into three groups of two each, into two groups of three each, and into five groups of four each is \frac{32!}{\left[ {\left( {2!} \right)^3 \left( {3!} \right)} \right]\left[ {\left( {3!} \right)^2 \left( {2!} \right)} \right]\left[ {\left( {4!} \right)^5 \left( {5!} \right)} \right]}
    Last edited by Plato; June 16th 2009 at 02:15 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. About minimal normal groups and subnormal groups
    Posted in the Advanced Algebra Forum
    Replies: 9
    Last Post: October 20th 2011, 02:53 PM
  2. Abelian groups and Homology groups
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: December 6th 2010, 02:44 PM
  3. Quotient Groups - Infinite Groups, finite orders
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: August 11th 2010, 08:07 AM
  4. free groups, finitely generated groups
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: May 23rd 2009, 04:31 AM
  5. Order of groups involving conjugates and abelian groups
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: February 5th 2009, 09:55 PM

Search Tags


/mathhelpforum @mathhelpforum