Combinatorics Problem: Choosing Couples in a Dance Class with 22 Students

  • Thread starter Thread starter arsato
  • Start date Start date
  • Tags Tags
    Combinatorics
Click For Summary
SUMMARY

The discussion centers on a combinatorial problem involving 22 students in a dance class, specifically selecting and pairing 5 men from a group of 12 and 5 women from a group of 10. The initial reasoning incorrectly calculated the number of combinations by multiplying available choices for each couple sequentially, leading to an inaccurate total of 410 combinations. The correct approach involves using combinations to first select 5 men from 12 and 5 women from 10, followed by calculating the distinct pairings of these selected individuals.

PREREQUISITES
  • Understanding of combinatorial mathematics
  • Familiarity with binomial coefficients
  • Knowledge of pairing and permutation concepts
  • Basic algebra for calculations
NEXT STEPS
  • Study the concept of binomial coefficients in combinatorics
  • Learn how to calculate combinations using the formula C(n, k)
  • Explore the principles of pairing and permutations in combinatorial problems
  • Practice similar combinatorial problems involving selections and pairings
USEFUL FOR

This discussion is beneficial for students studying combinatorics, educators teaching mathematical concepts, and anyone interested in solving complex selection and pairing problems in mathematics.

arsato
Messages
3
Reaction score
0

Homework Statement


A dance class consists of 22 students, of which 10 are women and 12 are men. If 5 men and 5 women are to be chosen and then paired off, how many results are possible?


Homework Equations





The Attempt at a Solution


We can say that for the first couple, there are a pool of 12 possible men to choose from and a pool of 10 possible women to choose from. So there are 12x10=120 possible couples. For the second couple, there are a pool of 11 men and a pool of 9 women. So 9x11=99. and so on until we get to the fifth couple (8x6=48). Then we add all the numbers together (120+99+80+63+48=410 possible 5 couple combinations). Is this the correct reasoning? I feel I'm missing something here.
 
Physics news on Phys.org
The reasoning is not correct. Consider a much simpler problem: you have two men, two women and you want to pair them into couples, how many ways can you do this? If we let the men be A and B, and the women be 1 and 2, then the only two distinct pairs of couples we have are:

(A,1) and (B,2)
(A,2) and (B,1)

so the answer for this is two. But by your previous reasoning we would conclude that the answer is 2*2 + 1*1 = 5.
 
Okay, so what would be a correct approach?
 
Given 10 men, how many ways can you pick 5 men out of it?
Given 12 women, how many ways can you pick 5 women out of it?
Given 5 men and 5 women, how many couples can you make?
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
5K