Suppose a company will select 3 people from a collection of 12 applicants to serve as a regional manager, an assistant regional manager, and an assistant to the regional manager. In how many ways can the selection be made?

Ok, so I'll start by doing my best to show where I am stuck by giving my best attempt at reasoning my way through this one:


At least one of the twelve applicants is a regional manager candidate. At least one of the twelve applicants is an assistant regional manager candidate. At least one of the twelve applicants is an applying for the assistant to the regional manager position.

So then what it is this question asking of me when it asks how many ways that a selection can be made? Any hints?

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    $\begingroup$ If all you would like is a hint: The question is asking how many ways in which the positions can be filled, eg: how many possibilities for the first position, how many for the second, etc.. Ultimately leading to the number of possible combinations of the applicants filling the positions. Look into permutations $\endgroup$ – frog1944 Mar 4 '16 at 4:49
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    $\begingroup$ Note: the word probability is used when talking about the likelihood of an event occurring and will always be between zero and one. This question asks for a count of how many arrangements are possible and will be a whole number (in this case) larger than one. Although we use counting methods developed for questions like these in several discrete probability settings, this specific question has nothing to do with probability per se and deserves a different tag. $\endgroup$ – JMoravitz Mar 4 '16 at 4:52

Hint: This question uses permutations.

Hover below to see answer.

There are 12 people and there are three distinguishable positions. You have 12 choices for the first position. Then you have 11 choices for the second position. Finally, once you have selected the other two, you have 10 people left for the last position. Hence, there are $12\cdot 11\cdot 10 = 1320$ ways to select the the committee/group.


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