ExplanationOne possible approach is to ask, “How many choices do I have for each of the class positions?” Begin by considering the president of the class. Since no one has been chosen yet, there
are 10 students from whom to choose. Then, for the vice president there are 9 options because now one student has already been chosen as president. Similarly, there are 8 choices for the secretary.
Using the fundamental counting principle, the total number of possible selections is (10)(9)(8) = 720.
Alternatively, use factorials. In this case, order matters because people are selected for specific positions. This problem is synonymous to asking, “How many different ways can you line up 3 students as first, second, and third from a class of 10?” The number of ways to arrange the entire class in line is 10!. However, the problem is only concerned with the first 3 students in line, so exclude rearrangements of the last 7.
The way in which these “non-chosen” 7 students can be ordered is 7!. Thus, the total number of arrangements for 3 students from a class of 10 is \(\frac{10!}{7!}= (10)(9)(8) =\) 720 choices.
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