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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 $29$...) – TMM Sep 20 '11 at 20:26
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 $(28)(27)(26)(1)$. – André Nicolas Sep 20 '11 at 20:34
  1. What you give, $8!$, is the number of different ways of ordering/ranking $8$ people. You aren't ranking them, you are selecting $8$ people from among $29$ available members. So this should be a "n-choose-k" answer; there are $29$ things to choose from, you have to pick $8$ things, so the answer is $\binom{29}{8} = \frac{29!}{21!8!}$.

    Alternatively: you have complete choice for the first person, $29$ possibilities; then $28$ left for the second, $27$ for the third, etc. So, a priori, you have $29\times28\times27\times26\times25\times24\times23\times 22=\frac{29!}{21!}$ ways. But the order in which you pick the team members is immaterial to who is on the team; each team can be ordered in $8!$ different ways, so you counted each tem $8!$ times. Dividing by $8!$ gives you the correct answer, $\frac{29!}{21!8!}$.

  2. This time, you have to select four people, but order matters because there are four different positions. So a slate can have anyone in the first position (29 possibilities), anyone left in the second position (28 possibilities), anyone else for the third (27), and any of the remaining 26 people for the final position. That gives $29\times 28\times 27\times 26$ possible slates.

  3. One of the positions is taken by Jake. That leaves three positions to be filled by the remaining 28 people in the club: proceeding as in 2, $28\times27\times26$ different ways of doing it.

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