1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Combinatorics groups

  1. Oct 1, 2012 #1
    Hi Everyone,

    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations
    c(n k)=n!/k!(n-k)!

    3. The attempt at a solution
    Included in Part A

    Sorry for this messy style.
  2. jcsd
  3. Oct 2, 2012 #2
    Re: Combinatorics

    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.
  4. Oct 2, 2012 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    Re: Combinatorics

    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.)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Combinatorics groups
  1. Combinatorics question (Replies: 2)

  2. Combinatorics problem (Replies: 2)

  3. Basic Combinatorics (Replies: 3)

  4. Combinatorics Question (Replies: 7)