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      <text>9.1.12 (Example of calculating combinations)</text>
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      <text><![CDATA[<h3>Example</h3>

<p>The School of Mathematics has 60 lecturers, and needs to choose 3 of them to teach Introduction to Linear Algebra. How many different lecturing teams are possible?</p>
<p>In this case the order of the people in the team does not matter. We can count the number of ways to pick 3 people in an ordered way:
\[
\dfrac{60!}{(60-3)!}=\dfrac{60\times59\times58\times57!}{57!}
\]
We just need to correct this for the fact that we don't care about the order -- so we divide by the number of ways to permute the 3 chosen lecturers:
\[
\dfrac{60!}{(60-3)!3!}=\dfrac{60\times59\times58\times57!}{57!3!}=\dfrac{60\times59\times58}{6}=10\times59\times58=34220
\]</p>
<br/>
<p>Alternatively we could have used the formula, but this reasoning helps illustrate where the formula comes from.</p>]]></text>
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