Combinatorics: Grouping 2n People into 2 Groups of n

In summary, there are (n-k)!/(n!)^2 different ways to make two groups out of a group of n people, and each of these ways has k members in it.
  • #1
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Hi Everyone,

Homework Statement


If we are asked the number of ways 2n people can be divided into 2 groups of n members,
can I first calculate the number of groups of n members that can be formed from 2n people and then calculate number of ways 2 groups can be selected from the number of groups formed above?

i.e. can I write:

number of groups possible=(2n)!/n!(2n-n)!=(2n)!/(n!)^2=k(suppose)

ways 2 group possible=k!/2!(k-2)!

And if I am further asked the same question with following addition:
if each department must choose a president and a vice president,
then can I multiply the number I got above by the number of ways a group can be formed with a vice president and a president. i.e. we will have n(n+1)/2 ways for selection of a president and (n(n+1)/2)^2 since we have two groups?
i.e. (k!/2!(k-2)!)*(n(n+1)/2)^2

Homework Equations


c(n k)=n!/k!(n-k)!

The Attempt at a Solution


Included in Part A

ThankYou.
Sorry for this messy style.
 
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  • #2


If you put n people in a group, then the other n people must of course go to the other group. But now suppose you swap the people between the groups. Does this count as a new way of making the groups, or is it identical to the one before?

So you already managed to figure out how many ways there are to make a group of n people out of a group of 2n people. Next you need to think very carefully about how many ways of choosing the 2 groups does each of these configurations have.
 
  • #3


Imagine making a list of the people, then putting "A" beside the names of the people to go in group "A", "B" beside the names of people to go in group "B". You now have a list of n "A"s and n "B"s. How many different permutations of that list are there? Finally, divide by 2 since it doesn't matter which group you call "A" and which group you call "B".

(I might have neglected to do that division by 2 if I hadn't first seen clamtrox's "But now suppose you swap the people between the groups. Does this count as a new way of making the groups, or is it identical to the one before?" I am NOT counting the as new ways of making two groups.)
 

FAQ: Combinatorics: Grouping 2n People into 2 Groups of n

1. What is combinatorics?

Combinatorics is a branch of mathematics that deals with counting and organizing combinations of objects or events.

2. What is the formula for grouping 2n people into 2 groups of n?

The formula for grouping 2n people into 2 groups of n is (2n)! / (n! * n!). This is known as the binomial coefficient or the choose function.

3. How do you solve a combinatorics problem involving grouping people?

To solve a combinatorics problem involving grouping people, you first need to determine the total number of people, the number of groups, and the size of each group. Then, use the formula (n1 + n2 + ... + nk)! / (n1! * n2! * ... * nk!) where n1, n2, ..., nk represent the number of people in each group.

4. Can a person be in both groups when using this formula?

No, the formula for grouping 2n people into 2 groups of n does not allow for a person to be in both groups. Each person can only be in one group.

5. What are some real-life applications of combinatorics?

Combinatorics has various real-life applications, such as in computer science for optimizing algorithms and data structures, in genetics for analyzing DNA sequences, in sports for creating tournament schedules, in cryptography for creating secure passwords, and in economics for studying market structures.

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