{"id":3537,"date":"2014-01-30T19:40:37","date_gmt":"2014-01-30T19:40:37","guid":{"rendered":"http:\/\/www.oracletutoring.ca\/blog\/?p=3537"},"modified":"2018-02-20T18:08:51","modified_gmt":"2018-02-20T18:08:51","slug":"math-a-counting-problem","status":"publish","type":"post","link":"https:\/\/www.oracletutoring.ca\/blog\/math-a-counting-problem\/","title":{"rendered":"Math:  A counting problem"},"content":{"rendered":"<h1>Tutoring math 12 and some university courses, you get asked about counting. The tutor opens the discussion with a couple of examples.<\/h1>\n<p>Questions that ask, &#8220;How many ways can people be lined up for a photograph&#8221; or &#8220;how many five-card hands have two aces&#8221; or &#8220;how many ways can you fill out a multiple choice test&#8221; are all counting problems.  <\/p>\n<p>When an item, once used, cannot be used again, we call that<em> without replacement.<\/em> Counting without replacement can often be done using permute or choose, but there are other options. \u00a0For a quick summary of permute and choose, see my article <a href=\"?p=1717\">here.<\/a><\/p>\n<p>When an item can be used over again, we call that <em>with replacement.<\/em>  A  question with replacement is often done using an exponential expression, as we shall see.<\/p>\n<p><strong>Example 1:  How many ways can you arrange the letters of the word radio?<\/strong><\/p>\n<p>Solution:  The answer is 5P5, or 120.  You use P &#8211; which means Permute &#8211; because the order of the letters is what matters, and because the process is without replacement:  that is, in a given arrangement, each letter can only appear once.<\/p>\n<p><strong>Example 2:  How many five card hands from a standard deck of 52 cards have exactly 2 aces?<\/strong><\/p>\n<p>Solution:  Here, the order doesn&#8217;t matter, but the process is still without replacement:  if you get the queen of hearts, you cannot draw it again in the same hand.  Without replacement, and where order doesn&#8217;t matter, points to Choose (aka Combination):  Specifically, (4C2)(48C3) or 103776.  The reasoning is as follows:  from the 4 aces, choose 2:  4C2.  Then, from the other 48 non-ace cards, choose 3:  48C3.  Multiply the two results together since they both happen yet are independent of each other.  Independent means that the choosing of the two aces has no influence on which non-ace cards will be chosen.<\/p>\n<p><strong>Example 3:  How many possible ways can a 20-question multiple choice test be filled out, if each question has five options?<br \/>\n<\/strong><br \/>\nSolution:  Order matters here, but it is <em>with replacement<\/em>:  for instance, you can choose answer A over and over again.  The number of ways you can fill out the test is 5^20.  The reasoning is that you can fill out the first question five possible ways, then the second five possible ways, and so on.  Your possibilities are (5)(5)(5)(5)&#8230;&#8230;(5)=5^20.  The number is absurdly large, so we&#8217;ll just refer to it as 5^20.<\/p>\n<p>I&#8217;ll say more about counting in future posts.  Hope your exams are going well:)<\/p>\n<p>Jack of <a href=\"https:\/\/www.oracletutoring.ca\">Oracle Tutoring by Jack and Diane,<\/a> Campbell River, BC.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Tutoring math 12 and some university courses, you get asked about counting. The tutor opens the discussion with a couple of examples. Questions that ask, &#8220;How many ways can people be lined up for a photograph&#8221; or &#8220;how many five-card &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/www.oracletutoring.ca\/blog\/math-a-counting-problem\/\"> <span class=\"screen-reader-text\">Math:  A counting problem<\/span> Read More &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[84,80,83,79,82,81],"class_list":["post-3537","post","type-post","status-publish","format-standard","hentry","category-math","tag-choose","tag-combination","tag-counting","tag-permutation","tag-with-replacement","tag-without-replacement"],"_links":{"self":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/3537","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/comments?post=3537"}],"version-history":[{"count":23,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/3537\/revisions"}],"predecessor-version":[{"id":30198,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/3537\/revisions\/30198"}],"wp:attachment":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/media?parent=3537"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/categories?post=3537"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/tags?post=3537"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}