{"id":1232,"date":"2017-12-19T02:37:41","date_gmt":"2017-12-19T02:37:41","guid":{"rendered":"https:\/\/kellyfish.me\/?p=1232"},"modified":"2019-02-17T18:42:15","modified_gmt":"2019-02-17T18:42:15","slug":"combinations","status":"publish","type":"post","link":"https:\/\/kellyfish.me\/index.php\/2017\/12\/19\/combinations\/","title":{"rendered":"Combinations"},"content":{"rendered":"

[latexpage]<\/span><\/p>\n

Fun Math for Girls
\n<\/span>By Kelly Tan<\/span><\/span><\/span><\/p>\n

\"\"<\/span><\/p>\n

Do you like taking school field trips? It would be fun, especially when you get to hang out with your friends. You could eat lunch together, go sightseeing with a group, and spend the day relaxing. <\/span><\/p>\n

Let’s say your mom is a chaperone, and can take 3 of your friends. Who do you invite to sit in your mom’s car? If you have 4 very good friends, Ashley (A), Brenda (B), Carol (C), and Diana (D), how many possible parties of 3 can you choose? Here are the possibilities:<\/span><\/p>\n

ABC
\n<\/span>ABD
\n<\/span>ACD
\n<\/span>BCD<\/span><\/span><\/p>\n

There are 4 different combinations. But what is the rule for counting here? Remember, from the previous permutation article, if you have 4 different colored beads (R,G,B,Y), and you want to make a 3 beaded necklace, the number of possibilities is $_4P_3=4$\u00b7$3$\u00b7$2=24$. But this is not the number of parties you can form from your 4 friends, because inviting ABC and BAC is considered the same. In other words, the order of the arrangement is not relevant here.<\/span><\/p>\n

For any 3 colors, says R, G, and B, that you pick, you have $3$\u00b7$2$\u00b7$1=6$ possible arrangements for a necklace:<\/span><\/p>\n

RGB
\n<\/span>RBG
\n<\/span>GRB
\n<\/span>GBR
\n<\/span>BRG
\n<\/span>BGR<\/span><\/span><\/p>\n

In this field trip problem, however, these 6 groups should all be considered as 1 combination. So, ABC, ACB, BAC, BCA, CAB, CBA are all counted as 1 combination. That will be true for any of the 3 friends group you chosen. So the number of combinations will be the number of permutations divided by $3$\u00b7$2$\u00b7$1=6$, that is $\\frac{24}{6}=4$\u00a0combinations.<\/span><\/p>\n

The notation that is used to indicate the combination is $_4C_3$<\/span>. So, as you just learned, the relationship between $_4C_3$<\/span>\u00a0<\/span>and $_4P_3$<\/span>\u00a0is: $_4C_3 = \\frac{_4P_3}{3!}$<\/span><\/span><\/p>\n

To try another example, suppose Emma(E) is another of your friends that you\u2019re thinking about inviting. Then how many parties could you form by picking 3 friends out of the 5?<\/span><\/p>\n

The answer is $_5C_3\u00a0= \\frac{_5P_3}{3!} = \\frac{5^.4^.3}{3!} = 10$ possible parties. This number is increasing quickly. The following is the list of all parties you can form by picking 3 out of 5 friends:<\/span><\/span><\/p>\n

ABC
\n<\/span>ABD
\n<\/span>ABE
\n<\/span>ACD
\n<\/span>ACE
\n<\/span>ADE
\n<\/span>BCD
\n<\/span>BCE
\n<\/span>BDE
\n<\/span>CDE<\/span><\/span><\/p>\n

In general, the formulas for permutation and combination, and their relationship, are summarized as follows:<\/span><\/p>\n

$_nP_r = \\frac{n!}{(n-r)!}$
\n<\/span> $_nC_r = \\frac{_nP_r}{r!} = \\frac{n!}{((n-r)!^.r!}$<\/span><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

[latexpage] Fun Math for Girls By Kelly Tan Do you like taking school field trips? It would be fun, especially when you get to hang out with your friends. You could eat lunch together, go sightseeing with a group, and spend the day relaxing. Let’s say your mom is a chaperone, and can take 3 […]<\/p>\n","protected":false},"author":1,"featured_media":1241,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[28],"tags":[],"class_list":["post-1232","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-articles"],"jetpack_featured_media_url":"https:\/\/i0.wp.com\/kellyfish.me\/wp-content\/uploads\/2018\/11\/b2878db66f14206c644d28e495ba6f15e3e3207a_1080.jpg?fit=1079%2C607&ssl=1","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/posts\/1232","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/comments?post=1232"}],"version-history":[{"count":12,"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/posts\/1232\/revisions"}],"predecessor-version":[{"id":1319,"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/posts\/1232\/revisions\/1319"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/media\/1241"}],"wp:attachment":[{"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/media?parent=1232"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/categories?post=1232"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kellyfish.me\/index.php\/wp-json\/wp\/v2\/tags?post=1232"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}