{"id":67975,"date":"2023-08-31T13:29:00","date_gmt":"2023-08-31T07:59:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=67975"},"modified":"2023-11-30T11:18:17","modified_gmt":"2023-11-30T05:48:17","slug":"rd-sharma-solutions-class-11-maths-chapter-16-exercise-16-4","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-16-exercise-16-4\/","title":{"rendered":"RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.4 (Updated 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-121909\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Class-11-Solutions-Chapter-16-Exercise-16.4.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.4\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Class-11-Solutions-Chapter-16-Exercise-16.4.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Class-11-Solutions-Chapter-16-Exercise-16.4-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.4<\/strong>: In Chapter 16 of Class 12 Mathematics, you will learn about factorial terms and symbols. Experts in the field simplify difficult problems into simple steps, and students can solve them with ease and precision. \u00a0<a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-16-permutations\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 16<\/a> Exercises 16.4 will help students acquire solid knowledge and master the subject.<\/p>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/cbse-class-11-maths-rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths<\/a>\u00a0<\/li>\n<\/ul>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" 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16.4\">Access RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.4<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-16-exercise-16-4\/#faq-rd-sharma-solutions-class-11-maths-chapter-16-exercise-164\" title=\"FAQ: RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.4\">FAQ: RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.4<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-16-exercise-16-4\/#is-rd-sharma-enough-for-class-12-maths\" title=\"Is RD Sharma enough for Class 12 Maths?\">Is RD Sharma enough for Class 12 Maths?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-16-exercise-16-4\/#what-are-the-benefits-of-studying-from-rd-sharma-solutions-class-12\" title=\"What are the benefits of studying from RD Sharma Solutions Class 12?\">What are the benefits of studying from RD Sharma Solutions Class 12?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-11-maths-chapter-16-exercise-164-pdf\"><\/span>Download RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.4 PDF:<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-16-Ex-16.4.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-16-Ex-16.4.pdf\">RD-Sharma-Solutions-Class-11-Maths-Chapter-16-Ex-16.4<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-solutions-class-11-maths-chapter-16-exercise-164\"><\/span>Access RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.4<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>1. In how many ways can the letters of the word \u2018FAILURE\u2019 be arranged so that the consonants may occupy only odd positions?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>The word \u2018FAILURE\u2019<\/p>\n<p>Number of vowels in word \u2018FAILURE\u2019 = 4(E, A, I, U)<\/p>\n<p>Number of consonants = 3(F, L, R)<\/p>\n<p>Let consonants be denoted by C<\/p>\n<p>Odd positions are 1, 3, 5 or 7<\/p>\n<p>The consonants can be arranged in these 4 odd places in\u00a0<sup>4<\/sup>P<sub>3<\/sub>\u00a0ways.<\/p>\n<p>The remaining 3 even places (2, 4, 6) are to be occupied by the 4 vowels. This can be done in\u00a0<sup>4<\/sup>P<sub>3<\/sub>\u00a0ways.<\/p>\n<p>So, the total number of words in which consonants occupy odd places =\u00a0<sup>4<\/sup>P<sub>3<\/sub>\u00a0\u00d7\u00a0<sup>4<\/sup>P<sub>3<\/sub><\/p>\n<p>By using the formula,<\/p>\n<p>P (n, r) = n!\/(n \u2013 r)!<\/p>\n<p>P (4, 3) \u00d7 P (4, 3) = 4!\/(4 \u2013 3)! \u00d7 4!\/(4 \u2013 3)!<\/p>\n<p>= 4 \u00d7 3 \u00d7 2 \u00d7 1 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1<\/p>\n<p>= 24 \u00d7 24<\/p>\n<p>= 576<\/p>\n<p>Hence, the number of arrangements so that the consonants occupy only odd positions is 576.<\/p>\n<p><strong>2. In how many ways can the letters of the word \u2018STRANGE\u2019 be arranged so that<br \/>(i) the vowels come together?<\/strong><\/p>\n<p><strong>(ii) the vowels never come together? And<\/strong><\/p>\n<p><strong>(iii) the vowels occupy only the odd places?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>The word \u2018STRANGE\u2019<\/p>\n<p>There are 7 letters in the word \u2018STRANGE\u2019, which includes 2 vowels (A,E) and 5 consonants (S,T,R,N,G).<\/p>\n<p><strong>(i)\u00a0<\/strong>The vowels come together.<\/p>\n<p>Considering 2 vowels as one letter so we will have 6 letters which can be arranged in\u00a0<sup>6<\/sup>P<sub>6<\/sub>\u00a0ways.<\/p>\n<p>(A,E) can be put together in\u00a0<sup>2<\/sup>P<sub>2<\/sub>\u00a0ways.<\/p>\n<p>Hence, the required number of words are<\/p>\n<p>By using the formula,<\/p>\n<p>P (n, r) = n!\/(n \u2013 r)!<\/p>\n<p>P (6, 6) \u00d7 P (2, 2) = 6!\/(6 \u2013 6)! \u00d7 2!\/(2 \u2013 2)!<\/p>\n<p>= 6! \u00d7 2!<\/p>\n<p>= 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 \u00d7 2 \u00d7 1<\/p>\n<p>= 720 \u00d7 2<\/p>\n<p>= 1440<\/p>\n<p>Hence, total number of arrangements in which vowels come together is 1440.<\/p>\n<p><strong>(ii)\u00a0<\/strong>The vowels never come together.<\/p>\n<p>The total number of letters in the word \u2018STRANGE\u2019 is\u00a0<sup>7<\/sup>P<sub>7<\/sub>\u00a0= 7! = 7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 = 5040<\/p>\n<p>So,<\/p>\n<p>Total number of words in which vowels never come together = total number of words \u2013 number of words in which vowels are always together<\/p>\n<p>= 5040 \u2013 1440<\/p>\n<p>= 3600<\/p>\n<p>Hence, the total number of arrangements in which vowels never come together is 3600.<\/p>\n<p><strong>(iii)\u00a0<\/strong>The vowels occupy only the odd places.<\/p>\n<p>There are 7 letters in the word \u2018STRANGE\u2019. Out of these letters (A,E) are the vowels.<\/p>\n<p>There are 4 odd places in the word \u2018STRANGE\u2019. The two vowels can be arranged in\u00a0<sup>4<\/sup>P<sub>2<\/sub>\u00a0ways.<\/p>\n<p>The remaining 5 consonants can be arranged among themselves in\u00a0<sup>5<\/sup>P<sub>5<\/sub>\u00a0ways.<\/p>\n<p>So, the total number of arrangements is<\/p>\n<p>By using the formula,<\/p>\n<p>P (n, r) = n!\/(n \u2013 r)!<\/p>\n<p>P (4, 2) \u00d7 P (5, 5) = 4!\/(4 \u2013 2)! \u00d7 5!\/(5 \u2013 5)!<\/p>\n<p>= 4!\/2! \u00d7 5!<\/p>\n<p>= (4\u00d73\u00d72!)\/2! \u00d7 5!<\/p>\n<p>= 4\u00d73 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1<\/p>\n<p>= 12 \u00d7 120<\/p>\n<p>= 1440<\/p>\n<p>Hence, the number of arrangements so that the vowels occupy only odd positions is 1440.<\/p>\n<p><strong>3. How many words can be formed from the letters of the word \u2018SUNDAY\u2019? How many of these begin with D?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>The word \u2018SUNDAY\u2019<\/p>\n<p>The total number of letters in the word \u2018SUNDAY\u2019 is 6.<\/p>\n<p>So, the number of arrangements of 6 things, taken all at a time, is\u00a0<sup>6<\/sup>P<sub>6<\/sub><\/p>\n<p>= 6! = 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 = 720<\/p>\n<p>Now, we shall find the number of words starting with D<\/p>\n<p>So let\u2019s fix the first position with the letter D, and then the remaining number of letters is 5.<\/p>\n<p>The number of arrangements of 5 things, taken all at a time, is\u00a0<sup>5<\/sup>P<sub>5<\/sub>\u00a0= 5! = 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 = 120<\/p>\n<p>Hence, the total number of words that can be made by letters of the word \u2018SUNDAY\u2019 is 720.<\/p>\n<p>The possible number of words using letters of \u2018SUNDAY\u2019 starting with \u2018D\u2019 is 120.<\/p>\n<p><strong>4. How many words can be formed out of the letters of the word, \u2018ORIENTAL,\u2019 so that the vowels always occupy odd places?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>The word \u2018ORIENTAL\u2019<\/p>\n<p>Number of vowels in the word \u2018ORIENTAL\u2019 = 4(O, I, E, A)<\/p>\n<p>Number of consonants in given word = 4(R, N, T, L)<\/p>\n<p>Odd positions are (1, 3, 5 or 7)<\/p>\n<p>Four vowels can be arranged in these 4 odd places in\u00a0<sup>4<\/sup>P<sub>4<\/sub>\u00a0ways.<\/p>\n<p>The remaining 4 even places (2,4,6,8) are to be occupied by the 4 consonants in\u00a0<sup>4<\/sup>P<sub>4<\/sub>\u00a0ways.<\/p>\n<p>So, by using the formula,<\/p>\n<p>P (n, r) = n!\/(n \u2013 r)!<\/p>\n<p>P (4, 4) \u00d7 P (4, 4) = 4!\/(4 \u2013 4)! \u00d7 4!\/(4 \u2013 4)!<\/p>\n<p>= 4! \u00d7 4!<\/p>\n<p>= 4 \u00d7 3 \u00d7 2 \u00d7 1 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1<\/p>\n<p>= 24 \u00d7 24<\/p>\n<p>= 576<\/p>\n<p>Hence, the number of arrangements so that the vowels occupy only odd positions is 576.<\/p>\n<p><strong>5. How many different words can be formed with the letters of the word \u2018SUNDAY\u2019? How many of the words begin with N? How many begin with N and end in Y?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>The word \u2018SUNDAY\u2019<\/p>\n<p>The total number of letters in the word \u2018SUNDAY\u2019 is 6.<\/p>\n<p>So, the number of arrangements of 6 things, taken all at a time, is\u00a0<sup>6<\/sup>P<sub>6<\/sub><\/p>\n<p>= 6! = 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 = 720<\/p>\n<p>Now, we shall find the number of words starting with N<\/p>\n<p>So let\u2019s fix the first position with the letter N, and then the remaining number of letters is 5.<\/p>\n<p>The number of arrangements of 5 things, taken all at a time, is\u00a0<sup>5<\/sup>P<sub>5<\/sub>\u00a0= 5! = 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 = 120<\/p>\n<p>Now, we need to find out a number of words starting with N and ending with Y<\/p>\n<p>So let\u2019s fix the first position with letters N and Y at the end, then the remaining number of letters is 4, which can be arranged in\u00a0<sup>4<\/sup>P<sub>4<\/sub>\u00a0ways. = 4! = 4 \u00d7 3 \u00d7 2 \u00d7 1 = 24<\/p>\n<p>Hence, the total number of words that can be made by letters of the word \u2018SUNDAY\u2019 is 720.<\/p>\n<p>The possible number of words using the letters of \u2018SUNDAY\u2019 starting with \u2018N\u2019 is 120.<\/p>\n<p>The possible number of words using the letters of \u2018SUNDAY\u2019 starting with \u2018N\u2019 and ending with \u2018Y\u2019 is 24.<\/p>\n<p><strong>6. How many different words can be formed from the letters of the word \u2018GANESHPURI\u2019? In how many of these words:<br \/>(i) the letter G always occupy the first place?<\/strong><\/p>\n<p><strong>(ii) the letter P and I respectively occupy the first and last place?<\/strong><\/p>\n<p><strong>(iii) Are the vowels\u00a0always together?<\/strong><\/p>\n<p><strong>(iv) the vowels always occupy even places?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>The word \u2018GANESHPURI\u2019<\/p>\n<p>There are 10 letters in the word \u2018GANESHPURI\u2019. The total number of words formed is\u00a0<sup>10<\/sup>P<sub>10<\/sub>\u00a0= 10!<\/p>\n<p><strong>(i)\u00a0<\/strong>The letter G always occupies the first place.<\/p>\n<p>If we fix the first position with the letter G, then the remaining number of letters is 9.<\/p>\n<p>The number of arrangements of 9 things, taken all at a time is\u00a0<sup>9<\/sup>P<sub>9<\/sub>\u00a0= 9! Ways.<\/p>\n<p>Hence, the possible number of words using letters of \u2018GANESHPURI\u2019 starting with \u2018G\u2019 is 9!<\/p>\n<p><strong>(ii)\u00a0<\/strong>The letters P and I respectively occupy the first and last place.<\/p>\n<p>If we fix the first position with letters P and I in the end, then the remaining number of letters is 8.<\/p>\n<p>The number of arrangements of 8 things, taken all at a time is\u00a0<sup>8<\/sup>P<sub>8<\/sub>\u00a0= 8! Ways.<\/p>\n<p>Hence, the possible number of words using letters of \u2018GANESHPURI\u2019 starting with \u2018P\u2019 and ending with \u2018I\u2019 is 8!<\/p>\n<p><strong>(iii)\u00a0<\/strong>The vowels are always together?<\/p>\n<p>There are 4 vowels and 6 consonants in the word \u2018GANESHPURI\u2019.<\/p>\n<p>Consider 4 (A,E,I,U) vowels as one letter, then the total number of letters is 7 (A,E,I,U, G, N, S, H , P, R)<\/p>\n<p>The number of arrangements of 7 things, taken all at a time, is\u00a0<sup>7<\/sup>P<sub>7<\/sub>\u00a0= 7! Ways.<\/p>\n<p>(A, E, I, U) can be put together in 4! Ways.<\/p>\n<p>Hence, the total number of arrangements in which vowels come together is 7! \u00d7 4!<\/p>\n<p><strong>(iv)\u00a0<\/strong>The vowels always occupy even places.<\/p>\n<p>Number of vowels in the word \u2018GANESHPURI\u2019 = 4(A, E, I, U)<\/p>\n<p>Number of consonants = 6(G, N, S, H, R, I)<\/p>\n<p>Even positions are 2, 4, 6, 8 or 10<\/p>\n<p>Now, we have to arrange 10 letters in a row such that vowels occupy even places. There are 5 even places (2, 4, 6, 8 or 10). 4 vowels can be arranged in these 5 even places in\u00a0<sup>5<\/sup>P<sub>4<\/sub>\u00a0ways.<\/p>\n<p>The remaining 5 odd places (1, 3, 5, 7, 9) are to be occupied by the 6 consonants in\u00a0<sup>6<\/sup>P<sub>5<\/sub>\u00a0ways.<\/p>\n<p>So, by using the formula,<\/p>\n<p>P (n, r) = n!\/(n \u2013 r)!<\/p>\n<p>P (5, 4) \u00d7 P (6, 5) = 5!\/(5 \u2013 4)! \u00d7 6!\/(6 \u2013 5)!<\/p>\n<p>= 5! \u00d7 6!<\/p>\n<p>Hence, the number of arrangements so that the vowels occupy only even positions is 5! \u00d7 6!<\/p>\n<p>We have included all the information regarding\u00a0<a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE\u00a0<\/a>RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.4. If you have any queries feel free to ask in the comment section.\u00a0<\/p>\n<h3><span class=\"ez-toc-section\" id=\"faq-rd-sharma-solutions-class-11-maths-chapter-16-exercise-164\"><\/span>FAQ: RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.4<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630396324073\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-rd-sharma-enough-for-class-12-maths\"><\/span>Is RD Sharma enough for Class 12 Maths?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>RD Sharma is a good book that gives you thousands of questions to practice.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630396362401\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-download-exercise-164-rd-sharma-solutions-class-11-maths-chapter-16-pdf-for-free\"><\/span>Can I download Exercise 16.4 RD Sharma Solutions Class 11 Maths Chapter 16 PDF for free?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can download RD Sharma Solutions Exercise 16.4 Class 11 Maths Chapter 16 PDF free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630396365223\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-open-rd-sharma-solutions-exercise-164-class-11-maths-chapter-16-pdf-on-my-smartphone\"><\/span>Can I open RD Sharma Solutions Exercise 16.4 Class 11 Maths Chapter 16 PDF on my smartphone?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, you can open RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.4 PDF on any device.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630396366225\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-studying-from-rd-sharma-solutions-class-12\"><\/span>What are the benefits of studying from RD Sharma Solutions Class 12?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>By practicing these solutions, students can earn higher academic grades. Our experts solve these solutions with utmost accuracy to help students in their studies.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.4: In Chapter 16 of Class 12 Mathematics, you will learn about factorial terms and symbols. Experts in the field simplify difficult problems into simple steps, and students can solve them with ease and precision. \u00a0RD Sharma Solutions Class 11 Maths Chapter 16 Exercises 16.4 will &#8230; <a title=\"RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.4 (Updated 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-16-exercise-16-4\/\" aria-label=\"More on RD Sharma Solutions Class 11 Maths Chapter 16 Exercise 16.4 (Updated 2024)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":121909,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[2917,73397,73717,73411,2985,73410],"tags":[3428,73334,73564],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67975"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/244"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=67975"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67975\/revisions"}],"predecessor-version":[{"id":514518,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67975\/revisions\/514518"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/121909"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=67975"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=67975"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=67975"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}