{"id":5907,"date":"2025-05-21T10:49:05","date_gmt":"2025-05-21T10:49:05","guid":{"rendered":"https:\/\/yaveni.com\/blog\/?p=5907"},"modified":"2025-05-21T10:49:07","modified_gmt":"2025-05-21T10:49:07","slug":"how-many-distinct-permutations-can-be-made-from-the-letters-of-the-word-philippines","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/how-many-distinct-permutations-can-be-made-from-the-letters-of-the-word-philippines\/","title":{"rendered":"How many distinct permutations can be made from the letters of the word PHILIPPINES"},"content":{"rendered":"\n<p>How many distinct permutations can be made from the letters of the word PHILIPPINES?<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>Let&#8217;s analyze the problem step-by-step.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Word:<\/strong> PHILIPPINES<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Count the total number of letters<\/h3>\n\n\n\n<p>The word <strong>PHILIPPINES<\/strong> has 10 letters in total.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Identify repeated letters and their frequencies<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>P appears 3 times<\/li>\n\n\n\n<li>I appears 3 times<\/li>\n\n\n\n<li>H appears 1 time<\/li>\n\n\n\n<li>L appears 1 time<\/li>\n\n\n\n<li>N appears 1 time<\/li>\n\n\n\n<li>E appears 1 time<\/li>\n\n\n\n<li>S appears 1 time<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Use the formula for permutations of multiset<\/h3>\n\n\n\n<p>The total number of <strong>distinct permutations<\/strong> of letters in a word with repeated letters is: n!n1!\u00d7n2!\u00d7\u22ef\\frac{n!}{n_1! \\times n_2! \\times \\cdots}<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>nn = total letters<\/li>\n\n\n\n<li>n1,n2,\u2026n_1, n_2, \\ldots = frequencies of each repeated letter<\/li>\n<\/ul>\n\n\n\n<p>For <strong>PHILIPPINES<\/strong>: n=10n = 10 nP=3,nI=3n_P = 3, \\quad n_I = 3<\/p>\n\n\n\n<p>All other letters appear once, so their factorials are 1 and don\u2019t affect the denominator.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Calculate factorials and apply the formula<\/h3>\n\n\n\n<p>Number&nbsp;of&nbsp;distinct&nbsp;permutations=10!3!\u00d73!\\text{Number of distinct permutations} = \\frac{10!}{3! \\times 3!}<\/p>\n\n\n\n<p>Calculate:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>10!=3,628,80010! = 3,628,800<\/li>\n\n\n\n<li>3!=63! = 6<\/li>\n<\/ul>\n\n\n\n<p>So, 3,628,8006\u00d76=3,628,80036=100,800\\frac{3,628,800}{6 \\times 6} = \\frac{3,628,800}{36} = 100,800<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final answer:<\/strong><\/h3>\n\n\n\n<p><strong>100,800 distinct permutations<\/strong> can be made from the letters of the word PHILIPPINES.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation (300 words):<\/h3>\n\n\n\n<p>When counting permutations of a word with repeated letters, it&#8217;s essential to account for indistinguishable arrangements caused by identical letters. The word &#8220;PHILIPPINES&#8221; contains 10 letters with two sets of repeated letters: three P&#8217;s and three I&#8217;s.<\/p>\n\n\n\n<p>If all letters were distinct, the total number of permutations would be 10!10!, which equals 3,628,800. This number represents every possible ordering of 10 unique letters.<\/p>\n\n\n\n<p>However, because three P&#8217;s are indistinguishable among themselves, swapping these P&#8217;s does not create new unique permutations. Similarly, the three I&#8217;s are identical and interchanging them also does not produce new distinct permutations. To correct for this overcounting, we divide by the factorial of the counts of each repeated letter.<\/p>\n\n\n\n<p>The number of ways to rearrange the three identical P&#8217;s among themselves is 3!=63! = 6, and the same applies for the three I&#8217;s. Dividing the total permutations by 3!\u00d73!3! \\times 3! removes duplicates that are identical due to these repetitions.<\/p>\n\n\n\n<p>Hence, the formula used is: 10!3!\u00d73!\\frac{10!}{3! \\times 3!}<\/p>\n\n\n\n<p>This yields: 3,628,8006\u00d76=100,800\\frac{3,628,800}{6 \\times 6} = 100,800<\/p>\n\n\n\n<p>Therefore, there are <strong>100,800<\/strong> unique ways to arrange the letters of &#8220;PHILIPPINES,&#8221; accounting properly for repeated letters. This approach is general and applies to any problem involving permutations of words with repeated letters, ensuring an accurate count of distinct permutations.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>How many distinct permutations can be made from the letters of the word PHILIPPINES? The correct answer and explanation is : Let&#8217;s analyze the problem step-by-step. Word: PHILIPPINES Step 1: Count the total number of letters The word PHILIPPINES has 10 letters in total. Step 2: Identify repeated letters and their frequencies Step 3: Use [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-5907","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/5907","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=5907"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/5907\/revisions"}],"predecessor-version":[{"id":5908,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/5907\/revisions\/5908"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=5907"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=5907"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=5907"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}