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Counting permutations

  1. Mar 13, 2015 #1
    1. The problem statement, all variables and given/known data
    Board of directors has 10 members. from the 10 members they will elect 5 officers. President, vice-pres, sec and treas

    A) From the 10 board members how many combinations of officers are there?

    B) If three board memebers are physicians, how many combinations have a physician as president?

    C) Exactly 1 physician as an officer?

    D) At least 1 physician as an officier?

    2. Relevant equationsm
    M×N

    3. The attempt at a solution
    A) 10×9×8×7×6, 10!/5!

    B) 3x9x8x7x6

    C) 3×7×6×5×4

    D) 3×9×8×7×6

    I just imagined plugging in people to the various office positions and used the simple M×N formula.
     
  2. jcsd
  3. Mar 13, 2015 #2

    Simon Bridge

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    Your approach is fine - for A you reduced the number each time because nobody can have two positions, and divided by 5! because the order is unimportant.

    For B: While there are three ways a physician can be president, you don't care which physician is president. Remember also that the order they get assigned still does not matter.
     
  4. Mar 13, 2015 #3
    So ur saying i should rethink solutions for b, c, d?
     
  5. Mar 13, 2015 #4

    Simon Bridge

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    It can help to check using letters and numbers:

    ABCDEFGHIJ are the members
    12345 are the jobs. 1 = president.

    ABC are doctors X = placeholder

    ADEFG
    BDEFG
    CDEFG
    ... each have a physician as president ... do we care if it's A or B or C who holds the post when it comes to counting unique combinations.
    I think that's a judgement call - also is ADEFG the same or different from ADFEG?
     
  6. Mar 13, 2015 #5

    haruspex

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    Yes, they count as different combinations here. I agree with the OP's answer for (b).
    The answer for (c) is not quite right. Chris, can you explian your reasoning there?
    The answer to (d) should be much higher than the answer to (b).
     
  7. Mar 13, 2015 #6
    It seems any good countable combination must have A, B or C in the president position however the other positions can permute all combinations a-j.... well minus whomever is president in that combination...


    c)

    I was thinking i have 2 pools of people. Physicians (3) and non-physicians(7)
    so i just plugged a physician in an office then used the 7 from the other pool to permute the other offices..
     
  8. Mar 13, 2015 #7

    Dick

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    For c) you have 3 different physicians and 5 different offices to fill in the one physician. Then populate the rest of the offices. For d) think about how many ways of populating use no physicians. The rest of them use at least one.
     
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