UMUC Stat Club must appoint a president, a vice president, and a treasurer. There are 8 qualified candidates. How many different ways can the officers be appointed?
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1$\begingroup$ Welcome to Math SE. In general, we ask for your attempted solution as well. Formatting tips here. $\endgroup$– Em.Commented Apr 16, 2017 at 4:07
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$\begingroup$ You don't need to know what either of those things are to do this problem (and you'll have better luck in life without them). Think it through. Must choose a president (8 choices) then a vice president (7 remaining choices) and so on. $\endgroup$– spaceisdarkgreenCommented Apr 16, 2017 at 4:07
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$\begingroup$ @spaceisdarkgreen , can I use 8C3 ? $\endgroup$– e4e5Commented Apr 16, 2017 at 4:15
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$\begingroup$ @e4e5 Sure you can just plug into a formula you don't understand that is probably relevant or at least related if you want to run a serious risk of getting the question wrong. Counting the possibilities yourself would be preferable $\endgroup$– spaceisdarkgreenCommented Apr 16, 2017 at 4:24
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1$\begingroup$ @e4e5 The reason it's not 8C3 as people have indicated below is cause the order of the choices (Pres, VP, and Treas) matters. The number of ways to pick a 3-person committee from $8$ people is $8C3.$ Notice why this is fewer. $\endgroup$– spaceisdarkgreenCommented Apr 16, 2017 at 4:28
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2 Answers
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The probability of this problem can be calculated through permutation. The number of ways of arrangement is $_{8}P_3 = \dfrac{8!}{5!} = 8\times7\times6 = 336$.
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$\begingroup$ Yu , are you sure it is permutation? No combination we use here? $\endgroup$– e4e5Commented Apr 16, 2017 at 4:22
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$\begingroup$ My last name is Yu by the way...Yes, I am pretty sure about this. It is a permutation since the assignment for president, vice president and treasurer is interchangable. If the problem asks for "picking a three-person block from 8 people", then it is combination, since "abc","bac" or "cab" represent the same block. $\endgroup$ Commented Apr 16, 2017 at 4:26
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$\begingroup$ @e4e5 This is a permutation problem since the positions of president, vice-president, and treasurer are distinct. If Angela, Barbara, and Celine are the officers, it matters whether Angela is the president, Barbara is the vice-president, and Celine is the treasure or Celine is the president, Barbara is the vice-president, and Angela is the treasurer. We use combinations when we make an unordered selection (choose a subset). We use permutations when order matters, as in this problem. $\endgroup$ Commented Apr 16, 2017 at 10:32
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If the picking or arranging different things at different positions required we use combination. But in this question no need of that. So you can use permutation in this case.
$$P^8_3= \frac {8!}{5!}$$
And other way as mentioned by one user in comment.
To pick president we have 8 options. Then for vice-president we have 7 options left. For treasurer we have 6 options. Then we have,
$$8 \times 7 \times 6$$
answered Apr 16, 2017 at 4:23
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$\begingroup$ Yes I thought we have 4 persons to select but then I see its 3 officers not 4. $\endgroup$ Commented Apr 16, 2017 at 4:26