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Set Theory - Counting - Binomial Coefficient - Factorials

  1. Jul 24, 2013 #1


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    1. The problem statement, all variables and given/known data

    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?

    2. Relevant 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!!!
  2. jcsd
  3. Jul 24, 2013 #2


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    Looks fine.
  4. Jul 24, 2013 #3


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    Thank you!
  5. Jul 25, 2013 #4


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    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!)
  6. Jul 25, 2013 #5


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    very clear thank you!

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