<p><a href="https://www.prepswift.com/quizzes/quiz/prepswift-introduction-to-combinatorics" target="_blank">Introduction to Combinatorics Exercise</a></p>
<p><strong>Note</strong>: at about 1:47 into the video, Greg says that </p>
<p>"number of permutations > number of combinations" </p>
<p>This should actually be</p>
<p>"number of permutations ≥ number of combinations" </p>
<p>He gives the example of choosing four people from five to make his point. The problem is that if we choose only one person, the number of permutations and combinations would be the same, that is, 5. </p><p>Let's do an <strong><span style="color:#8e44ad;">Introduction to Combinatorics</span></strong>. </p>
<p><strong><span style="color:#27ae60;">What is it? </span><span style="font-size:11px;"><em>From ChatGPT</em></span></strong></p>
<p>Combinatorics is the branch of math that studies how to count things, organize them, and figure out the number of possible arrangements (permutations) or combinations. It helps answer questions like:</p>
<ul>
<li>How many ways can I shuffle a deck of cards?</li>
<li>How many ways can a group of people sit in a row?</li>
<li>How many unique pizza combinations can I make with 5 toppings?</li>
</ul>
<p>It's all about understanding patterns, counting possibilities, and solving puzzles involving arrangements and choices.</p>
<p><strong><span style="color:#27ae60;">What are the two main components?</span></strong></p>
<p style="margin-left: 40px;"><u><strong>Permutations</strong></u>: The number of possible arrangements, where <u>order is important</u>.</p>
<p style="margin-left: 40px;"><strong><u>Combinations</u></strong>: The number of possible groupings, where order is <strong><span style="color:#e74c3c;"><u>not</u></span></strong> important.</p>
<p><strong><span style="color:#27ae60;">What do you mean order is important?</span></strong></p>
<p><em>Imagine we have the three letters $A$, $B$, and $C$</em></p>
<p style="margin-left: 40px;">In <strong><u>permutations</u></strong>, where order is important, note the following:</p>
<p>$$ABC \neq ACB \neq BAC \neq BCA \neq CBA \neq CAB$$</p>
<p style="margin-left: 80px;"><span style="color:#e74c3c;">Notice how the six different arrangements above are counted as distinct. These are six different cases.</span></p>
<p style="margin-left: 40px;">In <strong><u>combinations</u></strong>, however, the six cases are all one and the same.</p>
<p>$$ABC = ACB = BAC = BCA = CBA = CAB$$</p>
<p>As you can probably predict, in general, the number of permutations is greater than the number of combinations.</p>
<p><strong><span style="color:#27ae60;">Can you give a real example?</span></strong></p>
<p><em>Sure! Imagine we have five people: Albert, Bill, Cat, Dana, and Edward. </em></p>
<p style="margin-left: 40px;"><strong><u>Permutations</u></strong>: The five individuals run a race that awards different prizes for 1st, 2nd, and 3rd place. In how many different ways can these awards be given out?</p>
<p style="text-align: center;"><strong><span style="color:#27ae60;">Correct Answer $=60$</span></strong></p>
<p style="margin-left: 40px;"><strong><u>Combinations</u></strong>: Of the five individuals, a professor will randomly choose three to join him on a research expedition. In how many different ways can these three individuals be grouped?</p>
<p style="margin-left: 40px; text-align: center;"><strong><span style="color:#27ae60;">Correct Answer $=10$</span></strong></p>
<p><strong><span style="color:#e74c3c;">NOTE</span>: </strong>We will cover how to solve these types of problems in later mountain entries. For now, just notice that the number of permutations is generally higher than the number of combinations.</p>