divide into groups or arrange into groups

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June 15th 2009, 11:32 PM
s_ingram

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]12C2[/HTML] ways i.e. 12!/10!2!

The three people of the second group are chosen in [HTML]10C3[/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]12C6[/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]12C6[/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?
June 16th 2009, 03:04 AM
Plato

Quote:

Originally Posted by s_ingram

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]12C6[/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]12C6[/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.
June 16th 2009, 12:11 PM
s_ingram

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
June 16th 2009, 12:57 PM
Plato

Quote:

Originally Posted by s_ingram

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 $\frac{32!}{\left[ {\left( {2!} \right)^3 \left( {3!} \right)} \right]\left[ {\left( {3!} \right)^2 \left( {2!} \right)} \right]\left[ {\left( {4!} \right)^5 \left( {5!} \right)} \right]}$