# Combinatorics group question?

• Gridvvk
In summary, the question is how many ways can a group of 22 people be divided into 3 groups of 4 people and 2 groups of 5 people, without considering the order within each group. The answer is [(22 choose 4)(18 choose 4)(14 choose 4)(10 choose 5)(5 choose 5) ] / (3! * 2!) = 470531961900. This is equivalent to the formula of \frac{22!}{4!^35!^2}.\frac{1}{3!2!}.

#### Gridvvk

Let's say you have a group of 22 people, which you would like to break into 5 different groups -- 3 groups of 4 and 2 groups of 5. How many distinct ways can you form such groups?

I don't want to double count groups. Let's say I number the people from A - V. The group ABCDE and ACDBE should be considered the same group since they have the same elements.

I imagine the numerator is
[(22 choose 4)(18 choose 4)(14 choose 4)(10 choose 5)(5 choose 5) ]

I'm not sure what to divide by to get rid of counting extras.

Order the 5 bins, and order each position in each bin; there are then clearly 22! permutations. The items in the first 3 bins can be ordered 4! different ways and the items in the last 2 bins can be ordered 5! different ways, so there are a total of 4!4!4!5!5! orderings within bins.

It is not clear whether {ABCD, EFGH, ...} and {EFGH, ABCD, ...} are to be considered the same: if so, you need to reduce further by a factor of 3!2!.

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Observe that the familiar forumula for combinations ## \frac{n!}{k!(n-k)!} ## is derived in the same way where there are n objects and 2 bins containing k and n-k objects respectively, but when we use this we are not generally interested in the contents of the second bin.

Thank you. I wanted combinations not permutations though. So my first term would be [(22 choose 4)(18 choose 4)(14 choose 4)(10 choose 5)(5 choose 5) ]. If I understand correctly, then the inner ordering within the groups is already taken care of by the combination formula (dividing by k!). I did want to consider {ABCD, EFGH, ...} and {EFGH, ABCD, ...} to be the same, so I would divide by a factor of 3! 2! because I have 3 groups of size 4 and 2 groups of size 5. The 3! and 2! term are coming from the number of groups, not the sizes.

So the question:
How many ways to divide group of 22 people into 3 groups of 4 people and 2 groups of 5 people?

[(22 choose 4)(18 choose 4)(14 choose 4)(10 choose 5)(5 choose 5) ] / (3! * 2!) =
470531961900.

Gridvvk said:
Thank you. I wanted combinations not permutations though. So my first term would be [(22 choose 4)(18 choose 4)(14 choose 4)(10 choose 5)(5 choose 5) ]. If I understand correctly, then the inner ordering within the groups is already taken care of by the combination formula (dividing by k!). I did want to consider {ABCD, EFGH, ...} and {EFGH, ABCD, ...} to be the same, so I would divide by a factor of 3! 2! because I have 3 groups of size 4 and 2 groups of size 5. The 3! and 2! term are coming from the number of groups, not the sizes.

So the question:
How many ways to divide group of 22 people into 3 groups of 4 people and 2 groups of 5 people?

[(22 choose 4)(18 choose 4)(14 choose 4)(10 choose 5)(5 choose 5) ] / (3! * 2!) =
470531961900.

I have given you combinations, we get the same answer.

Note that a number of terms can be canceled from the expression you gave to arrive at the expression I gave: $$\frac{22!}{4!18!}.\frac{18!}{4!14!}.\frac{14!}{4!10!}.\frac{10!}{5!5!}.\frac{5!}{5!0!}.\frac{1}{3!2!} = \frac{22!}{4!^35!^2}.\frac{1}{3!2!}$$

1 person
Right, that eases the computation. Thanks a lot for the help!

## 1. What is combinatorics group theory?

Combinatorics group theory is a branch of mathematics that studies the properties of groups, which are mathematical structures that represent symmetries or transformations of objects. It involves the study of how to combine and manipulate elements of a group to produce new elements.

## 2. How is combinatorics group theory related to other branches of mathematics?

Combinatorics group theory is closely related to other areas of mathematics such as abstract algebra, number theory, and topology. It provides a powerful framework for understanding and solving problems in these fields.

## 3. What are some real-world applications of combinatorics group theory?

Combinatorics group theory has many applications in fields such as cryptography, coding theory, and physics. It is also used in computer science to analyze and optimize algorithms.

## 4. What are some important concepts in combinatorics group theory?

Some key concepts in combinatorics group theory include group operations such as multiplication and inversion, group structures such as subgroups and cosets, and group properties such as symmetry and closure.

There are many resources available for learning about combinatorics group theory, such as textbooks, online courses, and research papers. It is recommended to have a strong foundation in abstract algebra before diving into this subject. Additionally, attending conferences and seminars can provide valuable insights and opportunities for networking with other experts in the field.