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Help me solve this problem

  1. Oct 4, 2007 #1
    Given a club consisting of six men and seven women

    a. In how many ways can we select a committee of five persons?

    b. In how many ways can we select a committee of five persons with 2 ment and 3 women?
     
  2. jcsd
  3. Oct 4, 2007 #2

    arildno

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    This is a very difficult problem.
    None here at these forums will be able to solve it, unless YOU POST WHAT YOU HAVE DONE SO FAR!!
     
    Last edited by a moderator: Oct 4, 2007
  4. Oct 4, 2007 #3

    Avodyne

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    Actually, it's not that difficult. But arildno is right that you must show an attempt ...
     
  5. Oct 4, 2007 #4

    Dick

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    I guess people not showing work is making arildno crabby. Try to avoid this.
     
  6. Oct 5, 2007 #5

    HallsofIvy

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    Absolutely, you wouldn't like arildno when he is angry!
     
  7. Oct 5, 2007 #6
    I got it

    I went to a study help session last night and was able to get the answer. I would really like to have help in understanding combination, permutation, and other related problems.
     
  8. Oct 5, 2007 #7
    Here is what we came up with last night:

    Given a club consisting of six men and seven women
    a. In how many ways can we select a committee of five persons?

    This is a fundamental counting principle: 13 X 12 X 11 X 10 X 9 = 154,440

    I will find out today if I was right or not.

    b. In how many ways can we select a committee of 5 persons with two men and 3 women?

    This is 7 choose 3 times 6 choose 2 = 7! divided by 3! X 4! times 6! divided by 2! times 4!
    when you simplify I get = 525
     
  9. Oct 5, 2007 #8

    arildno

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    Completely correct!
    The binomial, [tex]\binom{7}{3}\equiv\frac{7!}{3!4!}[/tex] tells us that
    i) There are 7! permutations of 7 women. When these women are divided into two groups, (those 3 to be in the committee and those 4 who don't), then there are 3!*4! different permutations of a particular choice.

    Since permutations of a particular choice is irrelevant, we have to divide 7! with 3!4! to get the number of DISTINCT 3-committe groups of women. (i.e, 35)
    The number of distinct women groups of 3 is then multiplied with the number of distinct men groups of 2 (15), yielding a total of 525 possible committee choices.
     
    Last edited: Oct 5, 2007
  10. Oct 5, 2007 #9

    Dick

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    Part a) didn't go so well though. Just as arildno explained, the number of ways of picking 3 from 7 is [tex]\binom{7}{3}[/tex]. a) asks you to pick 5 from 13, why isn't the answer [tex]\binom{13}{5}[/tex]?
     
  11. Oct 5, 2007 #10

    arildno

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    Oops, you are right!
     
  12. Oct 20, 2007 #11
    it's not a difficult problem!!!!!
     
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