{"id":110,"date":"2019-09-23T02:11:20","date_gmt":"2019-09-23T02:11:20","guid":{"rendered":"http:\/\/blog.newtonchineseschool.org\/lilijia\/?p=110"},"modified":"2019-09-23T02:11:20","modified_gmt":"2019-09-23T02:11:20","slug":"aops_cp2-basic-counting-techniques","status":"publish","type":"post","link":"https:\/\/blog.newtonchineseschool.org\/lilijia\/2019\/09\/23\/aops_cp2-basic-counting-techniques\/","title":{"rendered":"AOPS_CP2: Basic Counting Techniques"},"content":{"rendered":"<p><strong>Lesson Review:<\/strong><\/p>\n<p>We introduced\u00a04\u00a0basic counting techniques today:<\/p>\n<p>1) <b>Casework<\/b>: to divide the problem into different <strong>subcases<\/strong>, count each, and add\u00a0the counts altogether. It helps break a counting problem into manageable pieces, but requires you to be organized and careful (do not omit any outcomes).<\/p>\n<p>2) <strong>Complementary Method<\/strong>: what we want = total = what we do not want. <strong>n(A) = n(U)-n(A&#8217;)<\/strong><\/p>\n<p>3)\u00a0<strong>Constructive Method<\/strong>: count how many ways we can construct a particular item.<\/p>\n<p>4)<strong> Counting with restrictions<\/strong>: it is generally best to deal with the most severe restrictions first.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Some tips:<\/strong><\/p>\n<p>1) independent choices: <strong>multiply<\/strong> the number of options at each step; exclusive options: <strong>add<\/strong> the number of options.<\/p>\n<p>2) Sometimes when you get stuck at complicated problems, try to <strong>start with some simple cases<\/strong>, and find a pattern, or hint, to help you solve the &#8220;big problem&#8221;.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Homework:<\/strong><\/p>\n<p>Page 34: Ex2.2.1, 2.2.3, 2.2.5* (*bonus, highly recommended)<br \/>\nPage 38: Ex 2.3.1, 2.3.3<br \/>\nPage 41: Ex 2.4.1, 2.4.3<br \/>\nPage 44: Ex 2.5.1, 2.5.4<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Lesson Review: We introduced\u00a04\u00a0basic counting techniques today: 1) Casework: to divide the problem into different subcases, count each, and add\u00a0the counts altogether. It helps break a counting problem into manageable pieces, but requires you to be organized and careful (do not omit any outcomes). 2) Complementary Method: what we want = total = what we [&hellip;]<\/p>\n","protected":false},"author":157,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"ngg_post_thumbnail":0},"categories":[1],"tags":[],"_links":{"self":[{"href":"https:\/\/blog.newtonchineseschool.org\/lilijia\/wp-json\/wp\/v2\/posts\/110"}],"collection":[{"href":"https:\/\/blog.newtonchineseschool.org\/lilijia\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.newtonchineseschool.org\/lilijia\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.newtonchineseschool.org\/lilijia\/wp-json\/wp\/v2\/users\/157"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.newtonchineseschool.org\/lilijia\/wp-json\/wp\/v2\/comments?post=110"}],"version-history":[{"count":1,"href":"https:\/\/blog.newtonchineseschool.org\/lilijia\/wp-json\/wp\/v2\/posts\/110\/revisions"}],"predecessor-version":[{"id":111,"href":"https:\/\/blog.newtonchineseschool.org\/lilijia\/wp-json\/wp\/v2\/posts\/110\/revisions\/111"}],"wp:attachment":[{"href":"https:\/\/blog.newtonchineseschool.org\/lilijia\/wp-json\/wp\/v2\/media?parent=110"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.newtonchineseschool.org\/lilijia\/wp-json\/wp\/v2\/categories?post=110"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.newtonchineseschool.org\/lilijia\/wp-json\/wp\/v2\/tags?post=110"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}