Set Theory - Counting - Binomial Coefficient - Factorials

reenmachine
Gold Member
Messages
514
Reaction score
9

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!
 
Physics news on Phys.org
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:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top