{"id":31188,"date":"2021-03-28T13:00:42","date_gmt":"2021-03-28T13:00:42","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=31188"},"modified":"2025-01-12T12:57:43","modified_gmt":"2025-01-12T12:57:43","slug":"combinatorics-problems-in-thermodynamics","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/en\/combinatorics-problems-in-thermodynamics\/","title":{"rendered":"Combinatorics Problems in Thermodynamics"},"content":{"rendered":"<style>\n    p, ul, ol {\n        text-align: justify;\n    }\n    h1, h2 {\n        text-align: center;\n    }\n<\/style>\n<h1>Combinatorics Problems in Thermodynamics<\/h1>\n<p style=\"text-align:center;\"><em>How many ways are there to organize a physical system composed of millions of elements? In this class, we will explore how mathematics allows us to answer questions like this in the context of thermodynamics, from the distribution of energy quanta in atomic systems to calculating possible configurations in large-scale systems. Using tools such as combinatorics, logarithms, and Stirling&#8217;s formula, we will delve into how to manage extraordinarily large numbers and solve seemingly insurmountable problems.<\/em><\/p>\n<p style=\"text-align:center;\"><strong>Learning Objectives:<\/strong><br \/>\nBy the end of this class, students will be able to:<\/p>\n<ol>\n<li><strong>Understand<\/strong> how combinatorics problems apply to the context of thermodynamics, specifically in organizing physical systems.<\/li>\n<li><strong>Calculate<\/strong> possible configurations of atomic systems using combinatorial numbers.<\/li>\n<li><strong>Apply<\/strong> Stirling&#8217;s formula to estimate the order of magnitude of complex configurations.<\/li>\n<\/ol>\n<p style=\"text-align:center;\"><u>TABLE OF CONTENTS<\/u>:<br \/>\n<a href=\"#1\"><strong>Combinatorics Problems<\/strong><\/a><br \/>\n<a href=\"#2\"><strong>Problems with Large Numbers<\/strong><\/a><br \/>\n<a href=\"#3\">Using Logarithms and Stirling\u2019s Formula to Estimate Orders of Magnitude<\/a><br \/>\n<a href=\"#4\"><em>Development via Simplified Approximation<\/em><\/a><br \/>\n<a href=\"#5\"><em>Development via Ordinary Approximation<\/em><\/a><br \/>\n<a href=\"#6\"><strong>Examples of Combinatorial and Order of Magnitude Calculations<\/strong><\/a><br \/>\n<a href=\"#7\">Case 1: Large Factorials<\/a><br \/>\n<a href=\"#8\">Case 2: Large Combinations<\/a>\n<\/p>\n<p>A common question in certain physical situations is: How many distinct ways can a given system be organized? These combinatorial problems frequently arise in thermodynamics. Although they may initially seem simple, they become complex when incorporating extremely large numbers, such as <strong><a href=\"http:\/\/toposuranos.com\/material\/mol-y-masa-molar\/\" rel=\"noopener\" target=\"_blank\">Avogadro&#8217;s number<\/a><\/strong> <span class=\"katex-eq\" data-katex-display=\"false\">N_A<\/span>, which exemplifies how overwhelming it can be to work with magnitudes of this scale.<\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Combinatorics Problems<\/h2>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/K7_De9wux4A?si=q93_T8xh15EdfJ6B\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/center><\/p>\n<p>To understand the magnitude of the problems involving combinatorics in thermodynamics, let us consider the following example:<\/p>\n<h4>Example: Combinations of Energy Quanta<\/h4>\n<p>Suppose we have a system composed of 10 atoms. Each atom can only store 1 or 0 units of energy, referred to as <strong>energy quanta<\/strong>. How many distinct ways can these quanta be distributed if we have (a) 10 energy quanta and (b) 5 energy quanta?<\/p>\n<h5>Solution<\/h5>\n<p>We represent the atoms as spaces available for storing an energy quantum. If a space is filled, it means the corresponding atom already has its quantum of energy.<\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-zJa7H7iMOGY\/YF3QhG94SoI\/AAAAAAAAEuI\/olfdSHfeJOgKea3vHXEAyr5QCJ1ZjmcpgCLcBGAsYHQ\/s839\/cuantodeenergia.PNG\" \n               class=\"aligncenter lazyload\" alt=\"Combinatorics problems in thermodynamics\" width=\"400\" height=\"200\" \/><\/center><\/p>\n<p>To count the ways in which <span class=\"katex-eq\" data-katex-display=\"false\">k<\/span> energy quanta can be distributed among <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> spaces, we use the combinatorial number:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\binom{n}{k}=\\dfrac{n!}{k!(n-k)!}<\/span><\/span><\/p>\n<p>This calculation gives us the number <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega<\/span> of possible states.<\/p>\n<p><strong>(a)<\/strong> If there are 10 quanta distributed among 10 spaces, there is only one way to do it. Thus, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Omega=1<\/span><\/span>:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\Omega = \\binom{10}{10}=\\dfrac{10!}{10!(10-10)!} = \\dfrac{10!}{10!0!} = 1 <\/span><\/span><\/p>\n<p><strong>(b)<\/strong> For 5 quanta distributed among 10 spaces, we calculate:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n\\Omega &amp;= \\displaystyle\\binom{10}{5} \\\\ \\\\\n\n&amp;=\\dfrac{10!}{5!(10-5)!} = \\dfrac{10!}{5!\\cdot 5!} \\\\ \\\\\n\n&amp;= \\dfrac{5! \\cdot 6\\cdot 7\\cdot 8 \\cdot 9\\cdot 10}{5! \\cdot 2\\cdot 3\\cdot 4\\cdot 5} \\\\ \\\\\n\n&amp;= \\dfrac{ 7\\cdot 8 \\cdot 9\\cdot 10}{ 4\\cdot 5} = 7\\cdot 2 \\cdot 9 \\cdot 2 = 252\n\n\\end{array}<\/span>\n<p>Therefore, there are 252 possible configurations.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Problems with Large Numbers<\/h2>\n<p>What we have analyzed so far is just the beginning. If we expand the system from case (b) to 100 atoms and 50 quanta, we obtain <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Omega \\approx 10^{28}<\/span><\/span>. Now, imagine performing the same calculation with a mole of atoms; the result would be inconceivable.<\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h3>Using Logarithms and Stirling\u2019s Formula to Estimate Orders of Magnitude<\/h3>\n<p>When we want to estimate a magnitude of the form <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Omega = \\binom{n}{k}<\/span><\/span> for large values of <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>, especially when <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=n\/2<\/span><\/span>, which is the case where maximum values are reached, it is useful to employ Stirling&#8217;s logarithmic approximation.<\/p>\n<p>To manage numbers of this magnitude, we can reformulate the calculations by taking logarithms, yielding:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\ln(\\Omega)=\\ln\\left(\\dfrac{n!}{k!(n-k)!}\\right)= \\ln(n!) - \\ln((n-k)!) - \\ln(k!)<\/span><\/span><\/p>\n<p>This expression can be refined using Stirling&#8217;s approximation for the logarithm of factorials. There are two possible versions of the approximation:<\/p>\n<ul>\n<li><strong>Ordinary approximation:<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln(n!) \\approx \\dfrac{1}{2}\\ln(2n\\pi) + n\\ln(n) - n<\/span><\/span> <\/li>\n<li><strong>Simplified approximation:<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln(n!) \\approx n\\ln(n) - n<\/span><\/span> <\/li>\n<\/ul>\n<p><a name=\"4\"><\/a><\/p>\n<h4>Development via Simplified Approximation<\/h4>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/fnt99FeeHPM?si=H6r-5GhO2yqaeaXd\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/center><\/p>\n<p>Using the simplified approximation, we obtain the following results:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n\\ln(\\Omega) &amp; \\approx n\\ln(n) - n - (n-k)\\ln(n-k) + (n-k) - k\\ln(k) + k \\\\ \\\\\n\n&amp;= n\\ln(n) - (n-k)\\ln(n-k) - k\\ln(k) \\\\ \\\\\n\n&amp;= n\\ln(n) - n\\ln(n-k) + k\\ln(n-k) - k\\ln(k) \\\\ \\\\\n\n&amp;= \\ln\\left[ \\left( \\dfrac{n}{n-k} \\right)^n \\right] + k\\ln\\left( \\dfrac{n-k}{k} \\right) \\\\ \\\\\n\n&amp;= \\ln\\left[ \\dfrac{1}{\\left(1 - \\dfrac{k}{n} \\right)^n} \\right] + k\\ln\\left( \\dfrac{n}{k} - 1 \\right)\n\n\\end{array}<\/span>\n<p>Since this approximation considers large values of <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>, we apply the relation:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\lim_{n\\to\\infty} \\left(1-\\dfrac{k}{n} \\right)^n = e^{-k} <\/span><\/span><\/p>\n<p>Thus:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\ln(\\Omega) \\approx \\ln(e^k) + k\\ln\\left( \\dfrac{n}{k} -1 \\right) = k + k\\ln\\left( \\dfrac{n}{k} -1 \\right) <\/span><\/span><\/p>\n<p>Finally, by applying a base change for logarithms, we obtain:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\log(\\Omega) = \\log(e)\\ln(\\Omega) \\approx k\\log(e)\\left[1 + \\ln\\left( \\dfrac{n}{k} - 1 \\right) \\right] <\/span><\/span><\/p>\n<p>This leads to the result:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\boxed{\\Omega \\approx 10^{k\\log(e)\\left[1 + \\ln\\left( \\dfrac{n}{k} - 1 \\right) \\right]}}<\/span><\/span><\/p>\n<p>Although this result does not provide the exact value of <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega<\/span>, it allows us to estimate the number of digits required to represent it, improving as <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> grows larger. With this method, it is sufficient to compute the exponent, which most calculators can handle.<\/p>\n<p>Additionally, this approach quickly estimates the maximum value of <span class=\"katex-eq\" data-katex-display=\"false\">\\Omega<\/span> for a large <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>. Considering the case where <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=n\/2<\/span><\/span>, we obtain:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\text{Max}\\left(\\Omega\\right) \\approx 10^{\\dfrac{n}{2}\\log(e)\\left[1 + \\ln\\left( \\dfrac{n}{n\/2} - 1 \\right) \\right]} = 10^{ n\\log(e)\/2 } <\/span><\/span><\/p>\n<p><a name=\"5\"><\/a><\/p>\n<h4>Development via Ordinary Approximation<\/h4>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/i3OO4lHV24Q?si=jEPCafXqgVxNYqH-\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/center><\/p>\n<p>While development using the ordinary approximation will yield a more precise result, it will involve additional calculations, leading to approximately equivalent results for large values of <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span>. The development of this approximation recycles several calculations already performed in the simplified approximation, as shown in the following reasoning:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rcl}\n\n\\ln(\\Omega) &amp; = &amp; \\ln\\left(\\dfrac{n!}{k!(n-k)!}\\right)= \\ln(n!) - \\ln((n-k)!) - \\ln(k!) \\\\ \\\\\n\n&amp; \\approx &amp; \\color{red}\\dfrac{1}{2}\\ln(2n\\pi)\\color{black} + n\\ln(n) - n \\\\ \\\\\n\n&amp; &amp; \\color{red}-\\dfrac{1}{2}\\ln(2(n-k)\\pi)\\color{black} - (n-k)\\ln(n-k) + (n-k) \\\\ \\\\\n\n&amp; &amp; \\color{red}-\\dfrac{1}{2}\\ln(2k\\pi)\\color{black} - k\\ln(k) + k\n\n\\end{array}<\/span>\n<p>The portion highlighted in red corresponds to the additional elements considered in the ordinary approximation, while everything else was already derived in the simplified approximation. Based on this, we have:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rcl}\n\n\\ln(\\Omega) &amp; \\approx &amp; \\color{red}\\dfrac{1}{2}\\ln\\left( \\dfrac{2n\\pi}{2(n-k)\\pi \\cdot 2k\\pi} \\right)\\color{black} + k + k\\ln\\left(\\dfrac{n}{k} - 1\\right) \\\\ \\\\\n\n&amp; = &amp;  k + k\\ln\\left(\\dfrac{n}{k} - 1\\right) - \\dfrac{1}{2}\\ln\\left(\\dfrac{2k\\pi(n-k)}{n}\\right)\n\n\\end{array}<\/span>\n<p>Then, using a logarithmic base change, we have:<\/p>\n<p style=\"text-align:center;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\log(\\Omega) = \\log(e)\\ln(\\Omega) \\approx \\log(e) \\left[ k + k\\ln\\left(\\dfrac{n}{k} - 1\\right) - \\dfrac{1}{2}\\ln\\left(\\dfrac{2k\\pi(n-k)}{n}\\right) \\right] <\/span><\/span><\/p>\n<p>Finally, taking the exponential with base 10, we get:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\Omega \\approx 10^{\\log(e) \\left[ k + k\\ln\\left(\\dfrac{n}{k} - 1\\right) - \\dfrac{1}{2}\\ln\\left(\\dfrac{2k\\pi(n-k)}{n}\\right) \\right]}\n\n<\/span>\n<p>Now, similarly to before, we can find the maximum value of this number by evaluating <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k=n\/2<\/span><\/span>, which in this case provides the following result:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rcl}\n\n\\text{Max}(\\Omega) &amp;\\approx &amp; 10^{\\log(e) \\left[ \\dfrac{n}{2} + \\dfrac{n}{2}\\ln\\left(\\dfrac{n}{(n\/2)} - 1\\right) - \\dfrac{1}{2}\\ln\\left(\\dfrac{2(n\/2)\\pi(n-n\/2)}{n}\\right) \\right]} \\\\ \\\\\n\n&amp; = &amp; 10^{\\log(e) \\left[\\dfrac{n}{2} - \\dfrac{1}{2}\\ln\\left(\\dfrac{n\\pi}{2} \\right) \\right]} = 10^{\\log(e)(n-\\ln(n\\pi\/2))\/2}\n\n\\end{array}<\/span>\n<p><a name=\"6\"><\/a><\/p>\n<h2>Examples of Combinatorial and Order of Magnitude Calculations<\/h2>\n<p><center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/M7NrtICrSzU?si=6UT34333Wp5YwFU1\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/center><\/p>\n<p><a name=\"7\"><\/a><\/p>\n<h3>Case 1: Large Factorials<\/h3>\n<p>Let us estimate the order of magnitude of <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left(10^{50}\\right)!<\/span><\/span>, i.e., the number of digits required to write this number.<\/p>\n<h5>Solution<\/h5>\n<p>To perform this calculation, we use Stirling&#8217;s formula as follows:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rl}\n\n\\ln\\left[ \\left(10^{50}\\right)! \\right] &amp;\\approx 10^{50}\\ln\\left(10^{50}\\right) - 10^{50}\\\\ \\\\\n\n&amp;= \\left[\\ln\\left(10^{50}\\right) -1\\right]10^{50}  \\\\ \\\\\n\n&amp;= \\left[50\\ln(10)-1 \\right]10^{50} \\\\ \\\\\n\n\\end{array}\n\n<\/span>\n<p>Next, we apply the base change for logarithms:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\ln\\left[ \\left(10^{50}\\right)! \\right] = \\dfrac{\\log\\left[\\left(10^{50}\\right)!\\right]}{\\log{e}}<\/span>\n<p>Thus:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\log\\left[ \\left(10^{50}\\right)! \\right] \\approx \\log(e)\\left[50\\ln(10)-1 \\right]10^{50}\n\n<\/span>\n<p>Finally, applying the base-10 exponential, we get:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\left(10^{50}\\right)!  \\approx  10^{\\log(e)\\left[50\\ln(10)-1 \\right]10^{50}} = 10^{49,5657 \\cdot 10^{50}}\n\n<\/span>\n<p>The exponent on the 10 represents the order of magnitude, providing an estimate of the number of digits in the number <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\left(10^{50}\\right)!<\/span><\/span>.<\/p>\n<p><a name=\"8\"><\/a><\/p>\n<h3>Case 2: Large Combinations<\/h3>\n<p>An average house has approximately 12 light switches, which can be either on or off. On average, each house accommodates 4 people. If a city has 5 million inhabitants, how many possible ways are there for half of the city\u2019s switches to be turned on?<\/p>\n<h5>Solution<\/h5>\n<p>The total number <span class=\"katex-eq\" data-katex-display=\"false\">n<\/span> of switches in the city is:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\begin{array}{rcl}\n\nn &amp;=&amp;\\dfrac{\\text{city inhabitants}}{\\text{people per house}} \\times \\text{switches per house} \\\\ \\\\\n\n&amp;=&amp; \\dfrac{5\\cdot 10^6}{4}\\cdot 12 = 15\\cdot 10^6\n\n\\end{array}\n\n<\/span>\n<p>The macrostate formed by all microstates where half of the switches are on coincides with the macrostate that has the largest number of possible configurations. Denoting this maximum number as <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Omega_{max}<\/span><\/span>, we can estimate it using each method:<\/p>\n<ul>\n<ol><strong>Ordinary Approximation:<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Omega_{max} = 10^{\\log(e)\\left[15\\cdot10^6 - \\ln\\left(15\\pi\\cdot10^6 \/ 2 \\right) \\right]\/2} \\approx 10^{6.514.413,542}<\/span><\/span><\/ol>\n<ol><strong>Simplified Approximation:<\/strong> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\Omega_{max} = 10^{\\log(e)\\left[15\\cdot10^6 \\right]\/2} \\approx 10^{6.514.417,229}<\/span><\/span><\/ol>\n<\/ul>\n<p>Although there is a difference of nearly 4 orders of magnitude between the two approximations (which may seem significant), this is actually negligible compared to the over 6.5 million orders of magnitude involved.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Combinatorics Problems in Thermodynamics How many ways are there to organize a physical system composed of millions of elements? In this class, we will explore how mathematics allows us to answer questions like this in the context of thermodynamics, from the distribution of energy quanta in atomic systems to calculating possible configurations in large-scale systems. [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":30534,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":27,"footnotes":""},"categories":[635,919],"tags":[],"class_list":["post-31188","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-physics","category-thermodynamics"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Combinatorics Problems in Thermodynamics - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"In this class, we will review an application of combinatorics and Stirling&#039;s formula in thermodynamics problems involving large numbers.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" 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