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Is my thinking correct? combinations of 2 types, wee

  1. Nov 10, 2006 #1
    Hello everyone, another example in the book i'm wanting to make sure i'm seeing how they got their answer correctly.

    It says:
    Supppose the group pof 12 consists of 5 men and 7 women.
    How many 5 person teams contain at most one man?

    Solution:
    The set of teams containing at most one man can be partitioned into the set that does not contain any men and the set that contains exactly one man. By the addition rule:

    [number of teams with at most one man] = [# of teams without any men] + [# of teams with one man]
    which equals:

    5 choose 0 * 7 choose 5 + 5 choose 1 * 7 choose 4 = 21 + 175 = 196.



    Okay I'm assuming they broke this down into a 2 step process.

    #1. choose the men
    #2. choose the women

    So for the # of teams without any men
    step 1:
    You have 5 men to pick from, but you don't want to choose any, so you say 5 choose 0.

    for step 2:
    you want all women but no men, you have 7 women to choose from but you only want a group of 5, so the max you can choose from the 7 women is 5. So 7 choose 5, now apply the muliplcation rule as you get:

    (5 choose 0) * (7 choose 5)

    now for the
    [# of teams with one man]
    step 1:
    choose men
    step 2:
    choose women

    Step 1: Choose 1 man out of 5, 5 choose 1.
    step 2: Choose 4 women out of the 7, so 7 choose 4, because you've already selected 1 man, so you have 4 slots open for the women.
    (5 choose 1) * (7 choose 4)

    Thanks!
     
  2. jcsd
  3. Nov 10, 2006 #2
    Looks perfect.
     
  4. Nov 10, 2006 #3
    sweet thanks again
     
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