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07 Jul 2025, 17:10
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We are given:
There are n married couples (so 2n people total).
We need to form a committee of n people.
We are to choose:
No married couples: No two people in the committee can be a married couple.
Only married couples: The committee of n people must be made only of married couples, so we're choosing n/2 couples.
Case 1: No married couples
We need to choose n people, but no two from the same couple.
From n couples, pick n people, one from each couple (i.e., 1 person per couple).
Each couple gives 2 choices (husband or wife): so for each of the n couples, we pick 1 person → 2 choices per couple.
So total ways =\(2^n\)
Case 2: Only married couples
We want to choose n people, and all must be in married pairs → choose n/2 couples from the n available.
Number of ways to choose n/2 couples from n couples = \(n!\) / \({(n/2)!}^2\)
There are n married couples (so 2n people total).
We need to form a committee of n people.
We are to choose:
No married couples: No two people in the committee can be a married couple.
Only married couples: The committee of n people must be made only of married couples, so we're choosing n/2 couples.
Case 1: No married couples
We need to choose n people, but no two from the same couple.
From n couples, pick n people, one from each couple (i.e., 1 person per couple).
Each couple gives 2 choices (husband or wife): so for each of the n couples, we pick 1 person → 2 choices per couple.
So total ways =\(2^n\)
Case 2: Only married couples
We want to choose n people, and all must be in married pairs → choose n/2 couples from the n available.
Number of ways to choose n/2 couples from n couples = \(n!\) / \({(n/2)!}^2\)
07 Jul 2025, 17:11
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no married couple in the committee
We must select 1 person from each of n couples notice that no two from the same couple).
Number of ways = 2^n (since each couple has 2 choices).
only married couple in committee
The committee must consist of n/2 full couples (since each couple has 2 people).
number of ways = C(n,n/2) = n!/(n/2)!^2 couples out of n
We must select 1 person from each of n couples notice that no two from the same couple).
Number of ways = 2^n (since each couple has 2 choices).
only married couple in committee
The committee must consist of n/2 full couples (since each couple has 2 people).
number of ways = C(n,n/2) = n!/(n/2)!^2 couples out of n
lvillalon
Joined: 29 Jun 2025
Last visit: 25 Aug 2025
Posts: 88
Given Kudos: 14
Location: Chile
Concentration: Operations, Entrepreneurship
Schools: HEC Sept '26 INSEAD Aug '26 IMD'26 Rotterdam'26 Booth '26 Sloan '26 Kellogg '26 Anderson '26
GPA: 3,3
WE:Consulting (Consulting)
Schools: HEC Sept '26 INSEAD Aug '26 IMD'26 Rotterdam'26 Booth '26 Sloan '26 Kellogg '26 Anderson '26
Posts: 88
07 Jul 2025, 19:13
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No married couples:
Instead of normal factorial, every time you choose an element, you got to discount 2 for the next multiplication, since it takes itself AND its partner away.
((2n)! / (2n! - 1)) / n!
(2n)!/(n!)2
Only married couples:
Takes the choosable elements from 2n to n, which have to be chosen for n/2 spots:
n!/(n/2)! = 2^n
Instead of normal factorial, every time you choose an element, you got to discount 2 for the next multiplication, since it takes itself AND its partner away.
((2n)! / (2n! - 1)) / n!
(2n)!/(n!)2
Only married couples:
Takes the choosable elements from 2n to n, which have to be chosen for n/2 spots:
n!/(n/2)! = 2^n
Bunuel
