{"id":67982,"date":"2023-08-12T13:50:00","date_gmt":"2023-08-12T08:20:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=67982"},"modified":"2023-11-10T10:42:32","modified_gmt":"2023-11-10T05:12:32","slug":"rd-sharma-solutions-class-11-maths-chapter-17-exercise-17-1","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-17-exercise-17-1\/","title":{"rendered":"RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.1 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-121917\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Class-11-Solutions-Chapter-17-Exercise-17.1.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.1\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Class-11-Solutions-Chapter-17-Exercise-17.1.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Class-11-Solutions-Chapter-17-Exercise-17.1-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.1<\/strong>: Previously we have studied the arrangement of a fixed number of objects acquiring some or all at the same time. Part or all of the elements, regardless of their arrangement, each of the different options is called a combination. Students who wish to learn by themselves and solve problems encountered in practice can use <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-17-combinations\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 17<\/a> Exercise 17.1\u00a0as reference material, which is the best resource available to any student. All solutions are created by experts in the field, taking into account the latest CBSE marking model. The exercise-based solutions for these topics are provided in PDF format, which can be easily downloaded by students.<\/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\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d7822ae5ab3\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" 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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-17-exercise-171-pdf\"><\/span>Download RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.1 PDF:<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div>\u00a0<\/div>\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-17-Ex-17.1.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-17-Ex-17.1.pdf\">RD-Sharma-Solutions-Class-11-Maths-Chapter-17-Ex-17.1<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-solutions-class-11-maths-chapter-17-exercise-171\"><\/span>Access RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.1<b><\/b><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>1. Evaluate the following:<\/strong><\/p>\n<p><strong>(i)\u00a0<sup>14<\/sup>C<sub>3<\/sub><\/strong><\/p>\n<p><strong>(ii)\u00a0<sup>12<\/sup>C<sub>10<\/sub><\/strong><\/p>\n<p><strong>(iii)\u00a0<sup>35<\/sup>C<sub>35<\/sub><\/strong><\/p>\n<p><strong>(iv)\u00a0<sup>n+1<\/sup>C<sub>n<\/sub><\/strong><\/p>\n<p><strong>(v)\u00a0<img title=\"RD Sharma Solutions for Class 11 Maths Chapter 17 \u2013 Combinations image - 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-17-combinations-image-1.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 17 \u2013 Combinations image - 1\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong><sup>14<\/sup>C<sub>3<\/sub><\/p>\n<p>Let us use the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p>So now, value of n = 14 and r = 3<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><sup>14<\/sup>C<sub>3<\/sub>\u00a0= 14! \/ 3!(14 \u2013 3)!<\/p>\n<p>= 14! \/ (3! 11!)<\/p>\n<p>= [14\u00d713\u00d712\u00d711!] \/ (3! 11!)<\/p>\n<p>= [14\u00d713\u00d712] \/ (3\u00d72)<\/p>\n<p>= 14\u00d713\u00d72<\/p>\n<p>= 364<\/p>\n<p><strong>(ii)\u00a0<\/strong><sup>12<\/sup>C<sub>10<\/sub><\/p>\n<p>Let us use the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p>So now, value of n = 12 and r = 10<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><sup>12<\/sup>C<sub>10<\/sub>\u00a0= 12! \/ 10!(12 \u2013 10)!<\/p>\n<p>= 12! \/ (10! 2!)<\/p>\n<p>= [12\u00d711\u00d710!] \/ (10! 2!)<\/p>\n<p>= [12\u00d711] \/ (2)<\/p>\n<p>= 6\u00d711<\/p>\n<p>= 66<\/p>\n<p><strong>(iii)\u00a0<\/strong><sup>35<\/sup>C<sub>35<\/sub><\/p>\n<p>Let us use the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p>So now, value of n = 35 and r = 35<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><sup>35<\/sup>C<sub>35<\/sub>\u00a0= 35! \/ 35!(35 \u2013 35)!<\/p>\n<p>= 35! \/ (35! 0!) [Since, 0! = 1]<\/p>\n<p>= 1<\/p>\n<p><strong>(iv)\u00a0<\/strong><sup>n+1<\/sup>C<sub>n<\/sub><\/p>\n<p>Let us use the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p>So now, value of n = n+1 and r = n<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><sup>n+1<\/sup>C<sub>n<\/sub>\u00a0= (n+1)! \/ n!(n+1 \u2013 n)!<\/p>\n<p>= (n+1)! \/ n!(1!)<\/p>\n<p>= (n + 1) \/ 1<\/p>\n<p>= n + 1<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 17 \u2013 Combinations image - 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-17-combinations-image-2.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 17 \u2013 Combinations image - 2\" \/><\/p>\n<p><strong>2. If\u00a0<sup>n<\/sup>C<sub>12<\/sub>\u00a0=\u00a0<sup>n<\/sup>C<sub>5<\/sub>, find the value of n.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that if\u00a0<sup>n<\/sup>C<sub>p<\/sub>\u00a0=\u00a0<sup>n<\/sup>C<sub>q<\/sub>, then one of the following conditions need to be satisfied:<\/p>\n<p>(i) p = q<\/p>\n<p>(ii) n = p + q<\/p>\n<p>So from the question\u00a0<sup>n<\/sup>C<sub>12<\/sub>\u00a0=\u00a0<sup>n<\/sup>C<sub>5<\/sub>, we can say that<\/p>\n<p>12 \u2260 5<\/p>\n<p>So, condition (ii) must be satisfied,<\/p>\n<p>n = 12 + 5<\/p>\n<p>n = 17<\/p>\n<p>\u2234\u00a0The value of n is 17.<\/p>\n<p><strong>3. If\u00a0<sup>n<\/sup>C<sub>4<\/sub>\u00a0=\u00a0<sup>n<\/sup>C<sub>6<\/sub>, find\u00a0<sup>12<\/sup>C<sub>n<\/sub>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that if\u00a0<sup>n<\/sup>C<sub>p<\/sub>\u00a0=\u00a0<sup>n<\/sup>C<sub>q<\/sub>, then one of the following conditions need to be satisfied:<\/p>\n<p>(i) p = q<\/p>\n<p>(ii) n = p + q<\/p>\n<p>So from the question\u00a0<sup>n<\/sup>C<sub>4<\/sub>\u00a0=\u00a0<sup>n<\/sup>C<sub>6<\/sub>, we can say that<\/p>\n<p>4 \u2260 6<\/p>\n<p>So, condition (ii) must be satisfied,<\/p>\n<p>n = 4 + 6<\/p>\n<p>n = 10<\/p>\n<p>Now, we need to find\u00a0<sup>12<\/sup>C<sub>n<\/sub>,<\/p>\n<p>We know the value of n so,\u00a0<sup>12<\/sup>C<sub>n<\/sub>\u00a0=\u00a0<sup>12<\/sup>C<sub>10<\/sub><\/p>\n<p>Let us use the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p>So now, value of n = 12 and r = 10<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><sup>12<\/sup>C<sub>10<\/sub>\u00a0= 12! \/ 10!(12 \u2013 10)!<\/p>\n<p>= 12! \/ (10! 2!)<\/p>\n<p>= [12\u00d711\u00d710!] \/ (10! 2!)<\/p>\n<p>= [12\u00d711] \/ (2)<\/p>\n<p>= 6\u00d711<\/p>\n<p>= 66<\/p>\n<p>\u2234\u00a0The value of\u00a0<sup>12<\/sup>C<sub>10<\/sub>\u00a0= 66.<\/p>\n<p><strong>4. If\u00a0<sup>n<\/sup>C<sub>10<\/sub>\u00a0=\u00a0<sup>n<\/sup>C<sub>12<\/sub>, find\u00a0<sup>23<\/sup>C<sub>n<\/sub>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that if\u00a0<sup>n<\/sup>C<sub>p<\/sub>\u00a0=\u00a0<sup>n<\/sup>C<sub>q<\/sub>, then one of the following conditions need to be satisfied:<\/p>\n<p>(i) p = q<\/p>\n<p>(ii) n = p + q<\/p>\n<p>So from the question\u00a0<sup>n<\/sup>C<sub>10<\/sub>\u00a0=\u00a0<sup>n<\/sup>C<sub>12<\/sub>, we can say that<\/p>\n<p>10 \u2260 12<\/p>\n<p>So, condition (ii) must be satisfied,<\/p>\n<p>n = 10 + 12<\/p>\n<p>n = 22<\/p>\n<p>Now, we need to find\u00a0<sup>23<\/sup>C<sub>n<\/sub>,<\/p>\n<p>We know the value of n so,\u00a0<sup>23<\/sup>C<sub>n<\/sub>\u00a0=\u00a0<sup>23<\/sup>C<sub>22<\/sub><\/p>\n<p>Let us use the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p>So now, value of n = 23 and r = 22<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><sup>23<\/sup>C<sub>22\u00a0<\/sub>= 23! \/ 22!(23 \u2013 22)!<\/p>\n<p>= 23! \/ (22! 1!)<\/p>\n<p>= [23\u00d722!] \/ (22!)<\/p>\n<p>= 23<\/p>\n<p>\u2234\u00a0The value of\u00a0<sup>23<\/sup>C<sub>22<\/sub>\u00a0= 23.<\/p>\n<p><strong>5. If\u00a0<sup>24<\/sup>C<sub>x<\/sub>\u00a0=\u00a0<sup>24<\/sup>C<sub>2x + 3<\/sub>, find x.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that if\u00a0<sup>n<\/sup>C<sub>p<\/sub>\u00a0=\u00a0<sup>n<\/sup>C<sub>q<\/sub>, then one of the following conditions need to be satisfied:<\/p>\n<p>(i) p = q<\/p>\n<p>(ii) n = p + q<\/p>\n<p>So from the question\u00a0<sup>24<\/sup>C<sub>x<\/sub>\u00a0=\u00a0<sup>24<\/sup>C<sub>2x + 3<\/sub>, we can say that<\/p>\n<p>Let us check for condition (i)<\/p>\n<p>x = 2x + 3<\/p>\n<p>2x \u2013 x = -3<\/p>\n<p>x = -3<\/p>\n<p>We know that for a combination\u00a0<sup>n<\/sup>C<sub>r<\/sub>, r\u22650, r should be a positive integer which is not satisfied here,<\/p>\n<p>So, condition (ii) must be satisfied,<\/p>\n<p>24 = x + 2x + 3<\/p>\n<p>3x = 21<\/p>\n<p>x = 21\/3<\/p>\n<p>x = 7<\/p>\n<p>\u2234\u00a0The value of x is 7.<\/p>\n<p><strong>6. If\u00a0<sup>18<\/sup>C<sub>x<\/sub>\u00a0=\u00a0<sup>18<\/sup>C<sub>x + 2<\/sub>, find x.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that if\u00a0<sup>n<\/sup>C<sub>p<\/sub>\u00a0=\u00a0<sup>n<\/sup>C<sub>q<\/sub>, then one of the following conditions need to be satisfied:<\/p>\n<p>(i) p = q<\/p>\n<p>(ii) n = p + q<\/p>\n<p>So from the question\u00a0<sup>18<\/sup>C<sub>x<\/sub>\u00a0=\u00a0<sup>18<\/sup>C<sub>x + 2<\/sub>, we can say that<\/p>\n<p>x \u2260 x + 2<\/p>\n<p>So, condition (ii) must be satisfied,<\/p>\n<p>18 = x + x + 2<\/p>\n<p>18 = 2x + 2<\/p>\n<p>2x = 18 \u2013 2<\/p>\n<p>2x = 16<\/p>\n<p>x = 16\/2<\/p>\n<p>= 8<\/p>\n<p>\u2234\u00a0The value of x is 8.<\/p>\n<p><strong>7. If\u00a0<sup>15<\/sup>C<sub>3r<\/sub>\u00a0=\u00a0<sup>15<\/sup>C<sub>r + 3<\/sub>, find r.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that if\u00a0<sup>n<\/sup>C<sub>p<\/sub>\u00a0=\u00a0<sup>n<\/sup>C<sub>q<\/sub>, then one of the following conditions need to be satisfied:<\/p>\n<p>(i) p = q<\/p>\n<p>(ii) n = p + q<\/p>\n<p>So from the question\u00a0<sup>15<\/sup>C<sub>3r<\/sub>\u00a0=\u00a0<sup>15<\/sup>C<sub>r + 3<\/sub>, we can say that<\/p>\n<p>Let us check for condition (i)<\/p>\n<p>3r = r + 3<\/p>\n<p>3r \u2013 r = 3<\/p>\n<p>2r = 3<\/p>\n<p>r = 3\/2<\/p>\n<p>We know that for a combination\u00a0<sup>n<\/sup>C<sub>r<\/sub>, r\u22650, r should be a positive integer which is not satisfied here,<\/p>\n<p>So, condition (ii) must be satisfied,<\/p>\n<p>15 = 3r + r + 3<\/p>\n<p>15 \u2013 3 = 4r<\/p>\n<p>4r = 12<\/p>\n<p>r = 12\/4<\/p>\n<p>= 3<\/p>\n<p>\u2234\u00a0The value of r is 3.<\/p>\n<p><strong>8. If\u00a0<sup>8<\/sup>C<sub>r<\/sub>\u00a0\u2013\u00a0<sup>7<\/sup>C<sub>3<\/sub>\u00a0=\u00a0<sup>7<\/sup>C<sub>2<\/sub>, find r.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>To find r, let us consider the given expression,<\/p>\n<p><sup>8<\/sup>C<sub>r<\/sub>\u00a0\u2013\u00a0<sup>7<\/sup>C<sub>3<\/sub>\u00a0=\u00a0<sup>7<\/sup>C<sub>2<\/sub><\/p>\n<p><sup>8<\/sup>C<sub>r<\/sub>\u00a0=\u00a0<sup>7<\/sup>C<sub>2<\/sub>\u00a0+\u00a0<sup>7<\/sup>C<sub>3<\/sub><\/p>\n<p>We know that\u00a0<sup>n<\/sup>C<sub>r<\/sub>\u00a0+\u00a0<sup>n<\/sup>C<sub>r + 1<\/sub>\u00a0=\u00a0<sup>n + 1<\/sup>C<sub>r + 1<\/sub><\/p>\n<p><sup>8<\/sup>C<sub>r<\/sub>\u00a0=\u00a0<sup>7 + 1<\/sup>C<sub>2 + 1<\/sub><\/p>\n<p><sup>8<\/sup>C<sub>r<\/sub>\u00a0=\u00a0<sup>8<\/sup>C<sub>3<\/sub><\/p>\n<p>Now, we know that if\u00a0<sup>n<\/sup>C<sub>p<\/sub>\u00a0=\u00a0<sup>n<\/sup>C<sub>q<\/sub>, then one of the following conditions need to be satisfied:<\/p>\n<p>(i) p = q<\/p>\n<p>(ii) n = p + q<\/p>\n<p>So from the question\u00a0<sup>8<\/sup>C<sub>r<\/sub>\u00a0=\u00a0<sup>8<\/sup>C<sub>3<\/sub>, we can say that<\/p>\n<p>Let us check for condition (i)<\/p>\n<p>r = 3<\/p>\n<p>Let us also check for condition (ii)<\/p>\n<p>8 = 3 + r<\/p>\n<p>r = 5<\/p>\n<p>\u2234\u00a0The values of \u2018r\u2019 are 3 and 5.<\/p>\n<p><strong>9. If\u00a0<sup>15<\/sup>C<sub>r<\/sub>:\u00a0<sup>15<\/sup>C<sub>r \u2013 1<\/sub>\u00a0= 11: 5, find r.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p><sup>15<\/sup>C<sub>r<\/sub>:\u00a0<sup>15<\/sup>C<sub>r \u2013 1<\/sub>\u00a0= 11: 5<\/p>\n<p><sup>15<\/sup>C<sub>r<\/sub>\u00a0\/\u00a0<sup>15<\/sup>C<sub>r \u2013 1<\/sub>\u00a0= 11 \/ 5<\/p>\n<p>Let us use the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 17 \u2013 Combinations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-17-combinations-image-3.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 17 \u2013 Combinations image - 3\" \/><\/p>\n<p>5(16 \u2013 r) = 11r<\/p>\n<p>80 \u2013 5r = 11r<\/p>\n<p>80 = 11r + 5r<\/p>\n<p>16r = 80<\/p>\n<p>r = 80\/16<\/p>\n<p>= 5<\/p>\n<p>\u2234\u00a0The value of r is 5.<\/p>\n<p><strong>10. If\u00a0<sup>n + 2<\/sup>C<sub>8<\/sub>:\u00a0<sup>n \u2013 2<\/sup>P<sub>4<\/sub>\u00a0= 57: 16, find n.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p><sup>n + 2<\/sup>C<sub>8<\/sub>:\u00a0<sup>n \u2013 2<\/sup>P<sub>4<\/sub>\u00a0= 57: 16<\/p>\n<p><sup>n + 2<\/sup>C<sub>8<\/sub>\u00a0\/\u00a0<sup>n \u2013 2<\/sup>P<sub>4<\/sub>\u00a0= 57 \/ 16<\/p>\n<p>Let us use the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 17 \u2013 Combinations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-17-combinations-image-4.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 17 \u2013 Combinations image - 4\" \/><\/p>\n<p>[(n+2)! (n-6)!] \/ [(n-6)! (n-2)! 8!] = 57\/16<\/p>\n<p>(n+2) (n+1) (n) (n-1) \/ 8! = 57\/16<\/p>\n<p>(n+2) (n+1) (n) (n-1) = (57\u00d78!) \/ 16<\/p>\n<p>(n+2) (n+1) (n) (n-1) = [19\u00d73 \u00d7 8\u00d77\u00d76\u00d75\u00d74\u00d73\u00d72\u00d71]\/16<\/p>\n<p>(n + 2) (n + 1) (n) (n \u2013 1) = 21 \u00d7 20 \u00d7 19 \u00d7 18<\/p>\n<p>Equating the corresponding terms on both sides, we get,<\/p>\n<p>n = 19<\/p>\n<p>\u2234\u00a0The value of n is 19.<\/p>\n<p>We have included all the information regarding\u00a0<a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.1. 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-17-exercise-171\"><\/span>FAQ: RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.1<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-1630397582628\" 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-1630397594918\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-open-rd-sharma-solutions-class-11-maths-chapter-17-exercise-171-pdf-on-my-smartphone\"><\/span>Can I open RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.1 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 17 Exercise 17.1 PDF on any device.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630397598623\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-download-the-rd-sharma-solutions-exercise-171-class-11-maths-chapter-17-pdf-for-free\"><\/span>Can I download the RD Sharma Solutions Exercise 17.1 Class 11 Maths Chapter 17 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 Class 11 Maths Chapter 17 Exercise 17.1 PDF free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630397601324\" 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 17 Exercise 17.1: Previously we have studied the arrangement of a fixed number of objects acquiring some or all at the same time. Part or all of the elements, regardless of their arrangement, each of the different options is called a combination. Students who wish to learn by &#8230; <a title=\"RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.1 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-17-exercise-17-1\/\" aria-label=\"More on RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.1 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":121917,"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\/67982"}],"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=67982"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67982\/revisions"}],"predecessor-version":[{"id":505696,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67982\/revisions\/505696"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/121917"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=67982"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=67982"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=67982"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}