Set Theory - Counting - Binomial Coefficient - Factorials

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Homework Help Overview

The problem involves selecting a committee from a department consisting of 5 men and 7 women, specifically choosing 3 men and 2 women. The discussion centers around the combinatorial methods to determine the number of ways to form this committee.

Discussion Character

  • Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss different methods for calculating the combinations of men and women, including using factorials and reasoning through choices without formulas. There is also a mention of the implications of gender representation in the committee selection.

Discussion Status

Several participants affirm the calculations presented, indicating that the approaches seem valid. There is a general agreement on the methods used, although the discussion also touches on social commentary regarding gender representation.

Contextual Notes

Some participants express concern about the gender dynamics in the committee selection process, noting the disparity in representation despite the larger number of women in the department.

reenmachine
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Homework Statement



A department consists of 5 men and 7 women.From this department you select a committee with 3 men and 2 women.In how many ways can you do this?

Homework Equations



Since the "overall set" (the entire department) is composed of both men and women and each has a specific number in the committee , I think I have to split it in half.

I pick 3 men out of 5 men and pick 2 women out of 7 women so:

##\binom{5}{3} \cdot \binom{7}{2} = \frac{5!}{3!(5-3)!} \cdot \frac{7!}{2!(7-2)!} = \frac{5!}{3!2!} \cdot \frac{7!}{2!5!} = \frac{120}{6 \cdot 2} \cdot \frac{5040}{2 \cdot 120} = \frac{120}{12} \cdot \frac{5040}{240} = 10 \cdot 21 = 210##

So 210 ways to form the committee of 3 men and 2 women from the department consisting of 5 men and 7 women.

thoughts on this?

thank you!
 
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Looks fine.
 
haruspex said:
Looks fine.

Thank you!
 
Without "formulas", you can argue that there are 5 choices for the first man, then 4 choices for the second, and 3 choices for the third- so 5(4)(3)= 60 ways to choose 3 out of 5- in that particular order. Since order does not matter divide by the 3!= 6 ways of ordering three people: 60/6= 10. For the women, there are 7 choices for the first woman then 6 choices for the second- so 7(6)= 42 ways to choose 2 out of 7- again in that particular order. Divide by the 2 different orders to get 42/2= 21 ways to do this. Overall there are (10)(21)= 210 ways to do this.

(Typical sexism- there are more women than men in the department but fewer on the commitee!)
 
HallsofIvy said:
Without "formulas", you can argue that there are 5 choices for the first man, then 4 choices for the second, and 3 choices for the third- so 5(4)(3)= 60 ways to choose 3 out of 5- in that particular order. Since order does not matter divide by the 3!= 6 ways of ordering three people: 60/6= 10. For the women, there are 7 choices for the first woman then 6 choices for the second- so 7(6)= 42 ways to choose 2 out of 7- again in that particular order. Divide by the 2 different orders to get 42/2= 21 ways to do this. Overall there are (10)(21)= 210 ways to do this.

(Typical sexism- there are more women than men in the department but fewer on the commitee!)

very clear thank you!

Yeah I noticed the sexism when I read it :smile:
 

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