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  <question type="description">
    <name>
      <text>9.1.2 (Example of calculating permutations)</text>
    </name>
    <questiontext format="moodle_auto_format">
      <text><![CDATA[<h3>Examples</h3>

<ol type="a">
<li>How many ways could the first 5 weeks of this course have been arranged?</li>
<li>A license plate begins with three letters. If the possible letters are A, B, C, D and E, how many different permutations of the letters can be made if no letter is used more than once?</li>
</ol>

<h3>Solution</h3>

<ol type="a">
<li>Recall that the number of ways of counting \(n\) objects, is given by \(n\) factorial. Here we have \(n=5\) so
\[
5!=5\times4\times3\times2\times1=120.
\]</li>
<li>For the first letter, there are 5 possible choices. After that letter is chosen, there are 4 possible choices. Finally, there are 3 possible choices for the last letter. Therefore the total number of choices is 
\[
5 × 4 × 3 = 60.
\]
Alternatively, we could use the permutation formula. Here we have \(n = 5\) and we need to choose \(r = 3\) letters. Thus the total number of permutations of the letters is 
\[
P(5,3)=\frac{}{}
\]</li>
</ol>]]></text>
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