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Permutations and combinations

  1. Jun 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Out of 7 women and 4 men you need to choose delegation.
    On how many ways you can choose delegation that consist of:
    a) five people - 3 women and 2 men
    b)any number of people, but with equal number of men and women
    c) five people with at least 3 women
    d) five peope where one member is decided to be women

    This is my attempt. I dont know d) part, help please :)

    [tex]a) \binom{7}{3}\binom{4}{2}=210\\ b)\binom{7}{1}\binom{4}{1} + \binom{7}{2}\binom{4}{2}+ \binom{7}{3}\binom{4}{3} + \binom{7}{4}\binom{4}{0}=329\\ c) \binom{7}{2}\binom{4}{3} + \binom{7}{3}\binom{4}{2} + \binom{7}{4}\binom{4}{1} + \binom{7}{5}\binom{4}{0}[/tex]
  2. jcsd
  3. Jun 19, 2015 #2
    Are you sure the first term in c) is correct? (At least 3 women...)
    And does d) say "one woman only", or what you've posted is the entire question part?
  4. Jun 19, 2015 #3
    You are right, its accident. first term shouldnt be there. ( expr. c)
    d) says that we have determined that one person is definitely women, other 4 can be both men and women.
    - So, first we can choose women in 7 ways. but how do i get other four ?

    By the way, a) and b) are correct?
  5. Jun 19, 2015 #4
    The number of women chosen for delegation can vary from 1 to 5 (since 5 people in total are to be selected and we need at least 1 woman to be chosen). The number of men chosen will then be ##5-n## where ##n## is the number of women that are chosen. So you will have something along the lines of "1 woman and 4 men" OR "2 women and 3 men" OR etc. Can you take it from here?
  6. Jun 19, 2015 #5
    I see that. I also have to do the same for men?

    One more thing. Do I then sum 1 women and 4 men + ... 4 men & 1 women?
  7. Jun 19, 2015 #6
    I don't get what you mean. "4 men and 1 woman" is the same thing as "1 woman and 4 men". You just have to cycle through different combinations by adding 1 to number of woman and reducing the number of men by 1 until you've exhausted all possibilities.
    No. Look above.
  8. Jun 19, 2015 #7
    Hmm, I tought something wrong.
    But i got it now.
    Thhanks !
  9. Jun 20, 2015 #8


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    There are only 4 men, right? So doesn't any selection of 5 people satisfy (d)?
  10. Jun 20, 2015 #9
    So basicly I have this:
    Five slots, where one is determined in advance to be women.
    Lets say its first slot.
    We can choose that women in 7 ways, so is this ok

    [tex]7[\binom{6}{1}\binom{4}{3} + \binom{6}{2}\binom{4}{2} + \binom{6}{3}\binom{4}{1}] +\binom{6}{5}[/tex]
  11. Jun 20, 2015 #10


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    The slots are not distinct, and neither are the women. According to (d), the only constraint is that at least one woman is chosen. Since 5 people are to be chosen and only 4 men are available, it is inevitable that at least one woman will be chosen.
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