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I'm reading a series of basic counting problems regarding a ten person club. I just want to make sure I'm sufficiently grasping the material and my problem solving methods are correct.

$1.$ In how many ways may a ten person club select a president and a secretary-treasurer from among its members?

For this problem, there are 10 ways to choose the president, and 9 ways to choose the secretary. In total, there are $10 \times 9 = 90$ ways to choose them.

$2.$ In how many ways may a ten person club select a two person executive committee from among its members?

This time, since we're taking 2 people from 10, I used ${10 \choose 2} = 45$ ways.

$3.$ In how many ways may a ten person club select a president and a two person executive advisory board from among its members (assuming that the president is not on the advisory board)?

I've already established from my first answer that there are 10 ways to choose the president. Then, I have to choose 2 people from a remaining group of 9. ${9 \choose 2} = 36$. Finally, I multiply the two results.$36\times10 = 360$ ways.

Any help would be appreciated, thank you.

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    $\begingroup$ It's correct :) $\endgroup$ – Angelo Mark Oct 31 '15 at 19:00
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    $\begingroup$ This is correct. $\endgroup$ – Antonios-Alexandros Robotis Oct 31 '15 at 19:02
  • $\begingroup$ Ok, great, thank you guys! $\endgroup$ – FeynmansBongos Oct 31 '15 at 19:04
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It is correct since you understand that the first answer requires a permutation since it is in a certain order, second requires combination because it is not chosen in any order, and the third is permutation and combination since one person is chosen in a specific order and the next two are chosen at random.

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All your answers are correct, and your reasoning is spot-on.

The only thing I'd like to point out is that for problem $1$, you can also use ${10 \choose 2} = 45$ to get the number of ways to pick the two people, and then there are two ways in which you can assign those two people to be president/secretary, and $45 \cdot 2 = 90$.

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