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<div id="main-column" class="span9"> <article id="permutations-combinations-sn" style="margin-top: 16px;">
<h1 class="section-title">Permutations & Combinations (SN)</h1>
<ul class="breadcrumb"><li><a title="Home" href="../../../index.html"><i class="fa fa-home"></i></a><span class="divider">/</span></li><li><span class="gray">1. Number & Algebra</span><span class="divider">/</span></li><li><span class="active">Permutations & Combinations (SN)</span></li></ul>
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<h3></h3><h3>Student Notes</h3><h4>I. Permutations</h4><p><img alt="" src="../../../ib/mathanalysis/analysis/number---algebra/permutations-combinations/perm-comb-img.jpg" style="float: left; width: 361px; height: 203px; margin-top: 1px; margin-bottom: 1px;"></p><div class="polaroid-right"><img height="136" src="../../../ib/mathanalysis/analysis/number---algebra/permutations-combinations/5-girls-run-img.jpg" style="margin: 8px 0px;" width="290"><div class="caption">Who comes in 1st, 2nd & 3rd place out of 5 runners?</div></div><hr><p>Andrea, Beatrice, Celia, Dinah and Eva are competing in an 800-metre run. Medals will be awarded for 1<sup>st</sup> place, 2<sup>nd</sup> place and 3<sup>rd</sup> place. How many different arrangements of the three medal winners are possible (assuming none of the top three finish with the exact same time)? Figure 1 below shows the 60 possible arrangements for 1<sup>st</sup>, 2<sup>nd</sup> and 3<sup>rd</sup> place for the five runners. The first column has all the possibilities for Andrea finishing first, the second column for Beatrice finishing first, etc. The arrangement of objects in a particular order is a called a <strong>permutation</strong>. For this 800-metre run, there are 60 permutations of 3 objects (1<sup>st</sup>, 2<sup>nd</sup> and 3<sup>rd</sup> places) from 5 objects (Andrea, Beatrice, Celia, Dinah & Eva). We need to develop a formula for computing permutations because trying to list all the permutations for a certain situation – as done in Figure 1 – is often impractical and/or too difficult.</p><div class="polaroid-center"><img src="../../../ib/mathanalysis/analysis/number---algebra/permutations-combinations/perm-list800-img.jpg" style="margin:8px 0"><div class="caption"> Figure 1</div></div><p>Before we develop a formula for computing the <strong>permutations of <em>r</em> objects selected from <em>n</em> objects</strong>, it is helpful to introduce a basic counting principle. The following example will illustrate what is often referred to as the <strong>multiplication principle</strong>.</p><p><strong><img alt="" src="../../../ib/mathanalysis/analysis/number---algebra/permutations-combinations/coffee-shop-img.jpg" style="margin-top: 1px; margin-bottom: 1px; float: left; width: 150px; height: 131px;">Example</strong></p><p>A coffee shop offers five different coffees: espresso (E), cappuccino (C), americano (A), latte (L) and macchiato (M); in one of three cup sizes: short (S), tall (T) or grande (G). How many different cups of coffee can be purchased?</p><p><strong>Solution</strong></p><p>There are two different actions: (1) choosing a type of coffee, and (2) choosing a cup size. There are five types of coffee from which to choose and three cup sizes from which to choose. Thus, there are <span class="math-tex">\(5 \times 3 = 15\)</span> ways to choose a cup of coffee. The <strong>tree diagram</strong> shows the 15 possible cups of coffee.</p><p style="text-align: center;"><img alt="" height="300" src="../../../ib/mathanalysis/analysis/number---algebra/permutations-combinations/tree-diag-15-coffees.jpg" style="margin-top: 1px; margin-bottom: 1px;" width="245"></p><div class="pinkBg"><p><strong>Multiplication Principle</strong></p><p>In a sequence of two actions (tasks, choices, events), if the first action can occur in <span class="math-tex">\({n_1}\)</span> ways and the second action can occur in <span class="math-tex">\({n_2}\)</span> ways, then the two actions can occur together in a total of <span class="math-tex">\({n_1} \times {n_2}\)</span> ways. In general, if there are <em><span class="math-tex">\(k\)</span></em> actions that can occur in <span class="math-tex">\({n_1},{n_2},{n_3}, \ldots ,{n_k}\)</span> ways, then the entire sequence of <em><span class="math-tex">\(k\)</span></em> actions can occur in <span class="math-tex">\({n_1} \times {n_2} \times {n_3} \times \cdots \times {n_k}\)</span> ways.</p></div><p>We can use the multiplication principle to quickly compute the number of permutations for 1<sup>st</sup>, 2<sup>nd</sup> and 3<sup>rd</sup> place for the five runners in the 800-metre run. We can consider each of the three medal winners from the five runners as an ‘action’ or ‘event’. (1) Firstly, there are five ‘choices’ for 1<sup>st</sup> place; (2) then there are four remaining ‘choices’ for 2<sup>nd</sup> place; (3) and, lastly, there are three remaining ‘choices’ for 3<sup>rd</sup> place. Hence, there are <span class="math-tex">\(5 \times 4 \times 3 = 60\)</span> permutations (ordered arrangements) for 1<sup>st</sup>, 2<sup>nd</sup> and 3<sup>rd</sup> place for the five runners. However, we wish to devise a general formula that we can apply in all situations where we wish to count the number of ordered arrangements, i.e. permutations. </p><p>Another helpful ingredient for building a formula for permutations is <strong>factorial notation</strong>. We notice that counting permutation always involves multiplying a succession of positive integers where each is one less than the previous integer; for example, <span class="math-tex">\(5 \times 4 \times 3\)</span> to count permutations for 1<sup>st</sup>, 2<sup>nd</sup> and 3<sup>rd</sup> place for five runners. To make counting calculations and many formulae easier to write, we use the ! symbol (factorial) to represent the multiplication of consecutive positive integers and is defined as follows: <span class="math-tex">\(n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times \cdots \times 3 \times 2 \times 1\)</span> where <em><span class="math-tex">\(n\)</span></em> is a nonnegative integer; and we define <span class="math-tex">\(0! = 1\)</span>. For example, <span class="math-tex">\(3! = 3 \cdot 2 \cdot 1 = 6\)</span> (<span class="math-tex">\(3!\)</span> is read as “3 factorial”, not “3 exclamation point”), and <span class="math-tex">\(7! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5040\)</span>.</p><p>Now we’re ready to compose a formula for computing permutations. For the people in the 800-metre run, note that the permutations of 3 runners (1<sup>st</sup>, 2<sup>nd</sup> and 3<sup>rd</sup> place) selected from 5 total runners is <span class="math-tex">\(5 \times 4 \times 3 = 60\)</span>. It is not <span class="math-tex">\(5!\)</span> but the calculation looks like <span class="math-tex">\(5!\)</span> at the start. However, consider that <span class="math-tex">\(5 \times 4 \times 3 = \frac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} = \frac{{5!}}{{2!}} = \frac{{5!}}{{\left( {5 - 3} \right)!}}\)</span>. This suggests the following formula.</p><div class="pinkBg"><h4><span style="font-size:12.0pt;line-height:115%;
font-family:"Times New Roman",serif;mso-fareast-font-family:Calibri;mso-fareast-theme-font:
minor-latin;mso-ansi-language:EN-GB;mso-fareast-language:EN-US;mso-bidi-language:
AR-SA">■</span> formula for permutations <span style="font-size:12.0pt;line-height:115%;
font-family:"Times New Roman",serif;mso-fareast-font-family:Calibri;mso-fareast-theme-font:
minor-latin;mso-ansi-language:EN-GB;mso-fareast-language:EN-US;mso-bidi-language:
AR-SA">■</span></h4><p>Let <span class="math-tex">\({}_n{{\textrm{P}}_r}\)</span> represent the <strong>number of</strong> <strong>permutations of <em>r</em> objects chosen from <em>n</em> objects</strong> which can also be stated as the <strong>number of</strong> <strong>permutations of <em>n</em> objects chosen <em>r</em> at a time</strong>.</p><p>Then, <span class="math-tex">\({}_n{{\textrm{P}}_r} = \frac{{n!}}{{\left( {n - r} \right)!}}\)</span> where <span class="math-tex">\(r \le n\)</span>.</p></div><p>Applying the formula to find the number of permutations of 5 objects chosen 3 at a time gives: <span class="math-tex">\({}_5{{\textrm{P}}_3} = \frac{{5!}}{{\left( {5 - 3} \right)!}} = \frac{{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}}{{2 \cdot 1}} = 5 \cdot 4 \cdot 3 = 60\)</span></p><h4>II. Combinations</h4><p>Let’s return to our 800-metre run with Andrea, Beatrice, Celia, Dinah and Eva. Let’s consider a different situation where we are not interested in the order (1<sup>st</sup>, 2<sup>nd</sup> and 3<sup>rd</sup>) of the top three finishers; but, instead, we are only interested in which of the five runners finish in the top three places. So, instead of counting <em>ordered</em> <em>arrangements</em> we wish to count <em>unordered selections</em>. That is, we want to count how many different subsets of three people can be selected from a group of five. An unordered selection of objects is called a <strong>combination</strong>. We can find the number of combinations of 3 objects (top three finishers) chosen from 5 objects (Andrea, Beatrice, Celia, Dinah & Eva) by using the list of permutations in Figure 1 and crossing out all the duplicate subsets of three. Since we are not interested in the order of the top three finishers than the result ABC is the same as ACB. In both cases, Andrea, Beatrice & Celia were the top three finishers and we are not interested in what order they finished; they are equivalent selections or subsets.</p><p>Figure 2 below shows that there are 10 combinations of 3 objects (top 3 finishers in any order) from 5 objects (Andrea, Beatrice, Celia, Dinah & Eva); or, in other words, there are 10 combinations of 5 objects chosen 3 at a time.</p><div class="polaroid-center"><img src="../../../ib/mathanalysis/analysis/number---algebra/permutations-combinations/perm-to-comb-img.jpg" style="margin:8px 0"><div class="caption"> Figure 2</div></div><p>Figure 3 below shows all the permutations of 5 objects chosen 3 at a time organized into 10 groups where each group contains 6 permutations that are identical if order is not important. Why are there 6 in each group? Because there are 6 permutations of 3 objects taken 3 at a time: <span class="math-tex">\({}_3{{\textrm{P}}_3} = \frac{{3!}}{{\left( {3 - 3} \right)!}} = \frac{{3!}}{{0!}} = \frac{{3!}}{1} = 3! = 6\)</span>. Note that <span class="math-tex">\({}_r{{\textrm{P}}_r} = r!\)</span></p><div class="polaroid-center"><img src="../../../ib/mathanalysis/analysis/number---algebra/permutations-combinations/perm-to-comb-hw-img.jpg" style="margin:8px 0"><div class="caption"> Figure 3</div></div><p>Therefore, one can determine the number of combinations of 5 objects chosen 3 at a time by taking the number of permutations of 5 objects chosen 3 at a time and dividing by 6 because there are 6 times more permutations than combinations (each group contains 6 identical combinations since they are unordered selections). In general, the number of combinations of <em><span class="math-tex">\(n\)</span></em> objects chosen <em><span class="math-tex">\(r\)</span></em> at a time is equal to the number of permutations for <em><span class="math-tex">\(n\)</span></em> choose <em><span class="math-tex">\(r\)</span></em> divided by <span class="math-tex">\(r!\)</span> Thus, we can use the formula we developed for counting permutations to write a formula for counting combinations.</p><div class="pinkBg"><h4><span style="font-size:12.0pt;line-height:115%;
font-family:"Times New Roman",serif;mso-fareast-font-family:Calibri;mso-fareast-theme-font:
minor-latin;mso-ansi-language:EN-GB;mso-fareast-language:EN-US;mso-bidi-language:
AR-SA">■</span> formula for combinations <span style="font-size:12.0pt;line-height:115%;
font-family:"Times New Roman",serif;mso-fareast-font-family:Calibri;mso-fareast-theme-font:
minor-latin;mso-ansi-language:EN-GB;mso-fareast-language:EN-US;mso-bidi-language:
AR-SA">■</span></h4><p>Let <span class="math-tex">\({}_n{{\textrm{C}}_r}\)</span> represent the <strong>number of</strong> <strong>combinations of <em>r</em> objects chosen from <em>n</em> objects</strong> which can also be stated as the <strong>number of</strong> <strong>combinations of <em>n</em> objects chosen <em>r</em> at a time</strong>.</p><p>Then, <span class="math-tex">\({}_n{{\textrm{C}}_r} = \frac{{n!}}{{r!\left( {n - r} \right)!}}\)</span> where <span class="math-tex">\(r \le n\)</span>.</p></div><p>Applying the formula to find the number of combinations of 5 objects chosen 3 at a time gives: <span class="math-tex">\({}_5{{\textrm{C}}_3} = \frac{{5!}}{{3!\left( {5 - 3} \right)!}} = \frac{{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}}{{\left( {3 \cdot 2 \cdot 1} \right)\left( {2 \cdot 1} \right)}} = \frac{{5 \cdot 4 \cdot 3}}{{3 \cdot 2 \cdot 1}} = \frac{{60}}{6} = 10\)</span></p><p>This confirms what Figures 2 & 3 show. For the 5 runners there are 10 different subsets of 3 runners that could be the first three to finish the run – and we are not interested in the order in which the first three finished.</p><hr><p style="text-align: center;"><img alt="" src="../../../ib/mathanalysis/analysis/number---algebra/permutations-combinations/perm-comb-table3-img.jpg" style="width: 600px; height: 309px;"></p><script>document.querySelectorAll('.tib-teacher-only').forEach(e => e.remove());</script>
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