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Hi Everyone,
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
c(n k)=n!/k!(n-k)!
Included in Part A
ThankYou.
Sorry for this messy style.
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.