Combinations Math Problem: Finding Possible Pairings with 10 Women and 12 Men

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Homework Help Overview

The problem involves forming pairs from a group of 10 women and 12 men, specifically creating 5 pairs where each pair consists of one man and one woman. The original poster attempts to determine the number of possible pairings and is exploring the combinatorial aspects of the selection and arrangement of these pairs.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of binomial coefficients to choose men and women, and the need to understand how to pair them once selected. There are questions about the reasoning behind multiplying by factorials and how to account for the arrangement of pairs.

Discussion Status

Participants are actively engaging with the problem, questioning the methods of counting arrangements and clarifying the reasoning behind their calculations. Some guidance has been provided regarding the arrangement of pairs, but there is no explicit consensus on the correct approach yet.

Contextual Notes

There seems to be some confusion regarding the correct method for counting the arrangements of pairs, with differing interpretations of how to approach the problem. The original poster and others are navigating through these assumptions and definitions.

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Homework Statement


there are 10 women and 12 men. 5 pairs are to be made, each pair having one man and one women. how many pairings are possible.

The Attempt at a Solution


first let's choose men and women from each set. there are ##{10 \choose 5}## women and ##{12 \choose 5}## men. now that we have these, what should i be thinking? i know the answer is ##{12 \choose 5} \times {10 \choose 5} \times 5!##. why multiply by the ##5!##?

thanks!
 
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Your binomial coefficients give the number of ways of choosing 5 women and 5 men. Now you have to figure out how many different pairs you can make with them.
 
In other words once you have chose the five men and five woman, how many ways can you order them man,woman,man,woman, etcetera...
 
LCKurtz said:
Your binomial coefficients give the number of ways of choosing 5 women and 5 men. Now you have to figure out how many different pairs you can make with them.

how many different pairs...this seems tough. well, I'm thinking of 5 positions. first, for the men there are ##5!## different ways to seat them. now for the women, there are also ##5!## ways to sit them. i think the answer is now ##5!+5!##. however, since women/man is the same as man/women, we divide by ##2!##?

is this correct?

thanks!
 
joshmccraney said:
how many different pairs...this seems tough. well, I'm thinking of 5 positions. first, for the men there are ##5!## different ways to seat them. now for the women, there are also ##5!## ways to sit them. i think the answer is now ##5!+5!##. however, since women/man is the same as man/women, we divide by ##2!##?

is this correct?

thanks!

No. Normally counting like that you would multiply the two numbers. But that isn't the correct analysis for this problem anyway. Look at it this way. All the men are seated and it doesn't matter what order. Now how many ways can you seat the ladies on their laps? There are your pairs.
 
Alright this is great! Makes perfect sense! Thanks!
 

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