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Thread: 2 part permutation calcuations

  1. #1
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    Question 2 part permutation calcuations

    A club has 8 members, namely A, B, C, D, E, F, G, and H.

    i.) The club needs to elect a president and secretary. How many ways can this be done if two members, G and H, can't stand each other and refuse to serve together in these positions?

    ii.) Suppose people A through E are women, the rest are men (F, G, H). If the choice of president and secretary is made uniformly at random, for example by drawing names out of a hat, what percentage of the time will both people selected be men?
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    Re: 2 part permutation calcuations

    Quote Originally Posted by azollner95 View Post
    A club has 8 members, namely A, B, C, D, E, F, G, and H.

    i.) The club needs to elect a president and secretary. How many ways can this be done if two members, G and H, can't stand each other and refuse to serve together in these positions?

    ii.) Suppose people A through E are women, the rest are men (F, G, H). If the choice of president and secretary is made uniformly at random, for example by drawing names out of a hat, what percentage of the time will both people selected be men?
    You must show us all some effort on your part.
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    Re: 2 part permutation calcuations

    Well for i) I got 48, because I assumed that any of the 8 members can be chosen, then it has to be 6 options left, because G and H cannot be together. So 8x6=48.

    for ii) I only got 6 because there are only 3 men (FGH) and if it is only men then the next option only leaves the other two guys so 3x2=6 ways.
    Last edited by azollner95; Apr 13th 2017 at 05:09 PM.
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    Re: 2 part permutation calcuations

    Quote Originally Posted by azollner95 View Post
    A club has 8 members, namely A, B, C, D, E, F, G, and H.
    i.) The club needs to elect a president and secretary. How many ways can this be done if two members, G and H, can't stand each other and refuse to serve together in these positions?
    ii.) Suppose people A through E are women, the rest are men (F, G, H). If the choice of president and secretary is made uniformly at random, for example by drawing names out of a hat, what percentage of the time will both people selected be men?
    i) There are three possible ways to do this:
    1. neither G nor H is chosen: $6\cdot 5$
    2. G is chosen but not H: $6\cdot 2$
    3. H is chosen but not G: $6\cdot 2$
    Now Add.

    This should give us the same answer as $8\cdot 7-2$ WHY?
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    Re: 2 part permutation calcuations

    2. (a) If G is chosen president any of the 6 people, not including H, can be chosen secretary. There are 6 ways to do this.
    (b) If H is chosen president any of the 6 people, not including G, can be chosen secretary. There are 6 ways to do this.
    (c) If any of the 6 people other than G and H is chosen president, any of the remaining 7 people can be chosen secretary. There are 7(6)= 42 ways to do this.

    There are 6+ 6+ 42= 54 ways to choose a president and secretary under these conditions.

    2) There are 8 people, 3 of them men. The probability that a president, chosen at random, is a man is 3/8. Given that that happens, there are 7 people from which to choose a secretary, 2 of them men. The probability a man is chosen secretary is 2/7. The probability two men are chosen, at random, is (3/8)(2/7)= 3/28.
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