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Can anyone take a look and let me know if I'm doing this right?

This year I am the advisor of a 29 member chapter of The Honor Society.

  1. How many different 8 member teams can I create.

    My Answer 8! 8x7x6x5x4x3x2x1=40320

  2. Elections must be 4 4 positions. All current members can be considered, how many different slates of officers can there be?

    Answer--I don't think the 29 students comes into play, I think it's just 4x3x2x1=24

  3. Jake really wants to be treasurer, and he is the only one interested. How many different slates of officers will there be if Jake is Treasurer?

    Answer: Again I don't think the 29 comes into play, but the first number is not a 4, but a 3. So 3x3x2x1=18

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    The first should be $\binom{29}{8} \gg 8!$: You can pick $1$ out of $29$ at first, then $1$ out of $28$, then $1$ out of $27$ etc. giving $29\cdot 28\cdot 27 \cdot \ldots \cdot 22$ options. However, you counted each team in $8!$ different orders, so you should divide by $8!$ giving $\binom{29}{8}$. Your other answers are wrong for similar reasons. (Note that for the first question, you can make way more different teams with $100000$ members than with $29$ members to choose from. Still your answer does not depend on the number $29$...)2011-09-20
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    I will assume that A for President, B for VP, C for Secretary, D for Treasurer is a different "slate" than C for Pres, B for VP, A for Secretary, and D for Treasurer. Then answer for different slates is $(29)(28)(27)(26)$. For the Pres. Choice can be any one of the 29 people, and for every choice for Pres. there are $28$ choices for VP, and so om. In the Jake case, same reasoning gives $(28)(27)(26)(1)$.2011-09-20

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