Combinatorics: Grouping 2n People into 2 Groups of n

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SUMMARY

The discussion focuses on calculating the number of ways to divide 2n people into 2 groups of n members. The formula used is (2n)!/(n!)^2 for determining the number of groups, followed by k!/2!(k-2)! for selecting the two groups. Additionally, when considering the selection of a president and vice president for each group, the total number of configurations is multiplied by (n(n+1)/2)^2. The importance of accounting for identical groupings when swapping members is emphasized, necessitating a division by 2 in the final calculation.

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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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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.
 


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.)
 

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