{"id":67983,"date":"2023-09-11T18:17:00","date_gmt":"2023-09-11T12:47:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=67983"},"modified":"2023-11-08T12:35:29","modified_gmt":"2023-11-08T07:05:29","slug":"rd-sharma-solutions-class-11-maths-chapter-17-exercise-17-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-17-exercise-17-2\/","title":{"rendered":"RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.2 (Updated 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-121921\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-17-Exercise-17.2.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-17-Exercise-17.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-17-Exercise-17.2-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.2<\/strong>: We previously studied the arrangement of a fixed number of objects, while acquiring some or all of them. Part or all of the elements, regardless of their arrangement, each different option 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 17.2 as reference material, which is the best resource that any student can use. All solutions are created by experts in the field and take into account the latest CBSE marking model. Exercise-based solutions for these topics are provided in PDF format, which students can download easily.<\/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-69da66d6bad4e\" 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\" width=\"20px\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-17-exercise-17-2\/#can-i-download-rd-sharma-solutions-exercise-172-class-11-maths-chapter-17-pdf-for-free\" title=\"Can I download RD Sharma Solutions Exercise 17.2 Class 11 Maths Chapter 17 PDF for free?\">Can I download RD Sharma Solutions Exercise 17.2 Class 11 Maths Chapter 17 PDF for free?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-17-exercise-17-2\/#can-i-open-rd-sharma-solutions-class-11-maths-chapter-17-exercise-172-pdf-on-my-smartphone\" title=\"Can I open RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.2 PDF on my smartphone?\">Can I open RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.2 PDF on my smartphone?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-17-exercise-17-2\/#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><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-11-maths-chapter-17-exercise-172-pdf\"><\/span>Download RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.2 PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>\u00a0<\/p>\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\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-17-Ex-17.2.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-17-Ex-17.2.pdf\">RD-Sharma-Solutions-Class-11-Maths-Chapter-17-Ex-17.2<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-rd-sharma-solutions-class-11-maths-chapter-17-exercise-172\"><\/span>Access RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>1. From a group of 15 cricket players, a team of 11 players is to be chosen. In how many ways can this be done?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>Number of players = 15<\/p>\n<p>Number of players to be selected = 11<\/p>\n<p>By using the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><sup>15<\/sup>C<sub>11<\/sub>\u00a0= 15! \/ 11! (15 \u2013 11)!<\/p>\n<p>= 15! \/ (11! 4!)<\/p>\n<p>= [15\u00d714\u00d713\u00d712\u00d711!] \/ (11! 4!)<\/p>\n<p>= [15\u00d714\u00d713\u00d712] \/ (4\u00d73\u00d72\u00d71)<\/p>\n<p>= 15\u00d77\u00d713<\/p>\n<p>= 1365<\/p>\n<p>\u2234\u00a0The total number of ways of choosing 11 players out of 15 is 1365 ways.<\/p>\n<p><strong>2. How many different boat parties of 8, consisting of 5 boys and 3 girls, can be made from 25 boys and 10 girls?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>Total boys are = 25<\/p>\n<p>Total girls are = 10<\/p>\n<p>Boat party of 8 to be made from 25 boys and 10 girls, by selecting 5 boys and 3 girls.<\/p>\n<p>So,<\/p>\n<p>By using the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><sup>25<\/sup>C<sub>5<\/sub>\u00a0\u00d7\u00a0<sup>10<\/sup>C<sub>3<\/sub>\u00a0= 25!\/5!(25 \u2013 5)! \u00d7 10!\/3!(10-3)!<\/p>\n<p>= 25! \/ (5! 20!) \u00d7 10!\/(3! 7!)<\/p>\n<p>= [25\u00d724\u00d723\u00d722\u00d721\u00d720!]\/(5! 20!) \u00d7 [10\u00d79\u00d78\u00d77!]\/(7! 3!)<\/p>\n<p>= [25\u00d724\u00d723\u00d722\u00d721]\/5! \u00d7 [10\u00d79\u00d78]\/(3!)<\/p>\n<p>= [25\u00d724\u00d723\u00d722\u00d721]\/(5\u00d74\u00d73\u00d72\u00d71) \u00d7 [10\u00d79\u00d78]\/(3\u00d72\u00d71)<\/p>\n<p>= 5\u00d72\u00d723\u00d711\u00d721 \u00d7 5\u00d73\u00d78<\/p>\n<p>= 53130 \u00d7 120<\/p>\n<p>= 6375600<\/p>\n<p>\u2234\u00a0The total number of different boat parties is 6375600 ways.<\/p>\n<p><strong>3. In how many ways can a student choose 5 courses out of 9 courses if 2 courses are compulsory for every student?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>The total number of courses is 9<\/p>\n<p>So out of 9 courses, 2 courses are compulsory. Students can choose from 7 (i.e., 5+2) courses only.<\/p>\n<p>That too out of 5 courses student has to choose, 2 courses are compulsory.<\/p>\n<p>So they have to choose 3 courses out of 7 courses.<\/p>\n<p>This can be done in\u00a0<sup>7<\/sup>C<sub>3<\/sub>\u00a0ways.<\/p>\n<p>By using the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><sup>7<\/sup>C<sub>3<\/sub>\u00a0= 7! \/ 3! (7 \u2013 3)!<\/p>\n<p>= 7! \/ (3! 4!)<\/p>\n<p>= [7\u00d76\u00d75\u00d74!] \/ (3! 4!)<\/p>\n<p>= [7\u00d76\u00d75] \/ (3\u00d72\u00d71)<\/p>\n<p>= 7\u00d75<\/p>\n<p>= 35<\/p>\n<p>\u2234 The total number of ways of choosing 5 subjects out of 9 subjects, of which 2 are compulsory, is 35 ways.<\/p>\n<p><strong>4. In how many ways can a football team of 11 players be selected from 16 players? How many of these will (i) Include 2 particular players? (ii) Exclude 2 particular players?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>Total number of players = 16<\/p>\n<p>Number of players to be selected = 11<\/p>\n<p>So, the combination is\u00a0<sup>16<\/sup>C<sub>11<\/sub><\/p>\n<p>By using the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><sup>16<\/sup>C<sub>11<\/sub>\u00a0= 16! \/ 11! (16 \u2013 11)!<\/p>\n<p>= 16! \/ (11! 5!)<\/p>\n<p>= [16\u00d715\u00d714\u00d713\u00d712\u00d711!] \/ (11! 5!)<\/p>\n<p>= [16\u00d715\u00d714\u00d713\u00d712] \/ (5\u00d74\u00d73\u00d72\u00d71)<\/p>\n<p>= 4\u00d77\u00d713\u00d712<\/p>\n<p>= 4368<\/p>\n<p>(i) Include 2 particular players?<\/p>\n<p>It is said that two players are always included.<\/p>\n<p>Now, we have to select 9 players out of the remaining 14 players as 2 players are already selected.<\/p>\n<p>Number of ways =\u00a0<sup>14<\/sup>C<sub>9<\/sub><\/p>\n<p><sup>14<\/sup>C<sub>9<\/sub>\u00a0= 14! \/ 9! (14 \u2013 9)!<\/p>\n<p>= 14! \/ (9! 5!)<\/p>\n<p>= [14\u00d713\u00d712\u00d711\u00d710\u00d79!] \/ (9! 5!)<\/p>\n<p>= [14\u00d713\u00d712\u00d711\u00d710] \/ (5\u00d74\u00d73\u00d72\u00d71)<\/p>\n<p>= 7\u00d713\u00d711\u00d72<\/p>\n<p>= 2002<\/p>\n<p>(ii) Exclude 2 particular players?<\/p>\n<p>It is said that two players are always excluded.<\/p>\n<p>Now, we have to select 11 players out of the remaining 14 players as 2 players are already removed.<\/p>\n<p>Number of ways =\u00a0<sup>14<\/sup>C<sub>9<\/sub><\/p>\n<p><sup>14<\/sup>C<sub>11<\/sub>\u00a0= 14! \/ 11! (14 \u2013 11)!<\/p>\n<p>= 14! \/ (11! 3!)<\/p>\n<p>= [14\u00d713\u00d712\u00d711!] \/ (11! 3!)<\/p>\n<p>= [14\u00d713\u00d712] \/ (3\u00d72\u00d71)<\/p>\n<p>= 14\u00d713\u00d72<\/p>\n<p>= 364<\/p>\n<p>\u2234\u00a0The required no. of ways are 4368, 2002, 364.<\/p>\n<p><strong>5. There are 10 professors and 20 students out of whom a committee of 2 professors and 3 students is to be formed. Find the number of ways in which this can be done. Further, find in how many of these committees:<br \/>(i) a particular professor is included.<\/strong><\/p>\n<p><strong>(ii) a particular student is included.<\/strong><\/p>\n<p><strong>(iii) a particular student is excluded.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>Total number of professors = 10<\/p>\n<p>Total number of students = 20<\/p>\n<p>Number of ways = (choosing 2 professors out of 10 professors) \u00d7 (choosing 3 students out of 20 students)<\/p>\n<p>= (<sup>10<\/sup>C<sub>2<\/sub>) \u00d7 (<sup>20<\/sup>C<sub>3<\/sub>)<\/p>\n<p>By using the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><sup>10<\/sup>C<sub>2<\/sub>\u00a0\u00d7\u00a0<sup>20<\/sup>C<sub>3<\/sub>\u00a0= 10!\/2!(10 \u2013 2)! \u00d7 20!\/3!(20-3)!<\/p>\n<p>= 10!\/(2! 8!) \u00d7 20!\/(3! 17!)<\/p>\n<p>= [10\u00d79\u00d78!]\/(2! 8!) \u00d7 [20\u00d719\u00d718\u00d717!]\/(17! 3!)<\/p>\n<p>= [10\u00d79]\/2! \u00d7 [20\u00d719\u00d718]\/(3!)<\/p>\n<p>= [10\u00d79]\/(2\u00d71) \u00d7 [20\u00d719\u00d718]\/(3\u00d72\u00d71)<\/p>\n<p>= 5\u00d79 \u00d7 10\u00d719\u00d76<\/p>\n<p>= 45 \u00d7 1140<\/p>\n<p>= 51300 ways<\/p>\n<p><strong>(i)\u00a0<\/strong>a particular professor is included.<\/p>\n<p>Number of ways = (choosing 1 professor out of 9 professors) \u00d7 (choosing 3 students out of 20 students)<\/p>\n<p>=\u00a0<sup>9<\/sup>C<sub>1<\/sub>\u00a0\u00d7\u00a0<sup>20<\/sup>C<sub>3<\/sub><\/p>\n<p>By using the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><sup>9<\/sup>C<sub>1<\/sub>\u00a0\u00d7\u00a0<sup>20<\/sup>C<sub>3\u00a0<\/sub>= 9!\/1!(9 \u2013 1)! \u00d7 20!\/3!(20-3)!<\/p>\n<p>= 9!\/(1! 8!) \u00d7 20!\/(3! 17!)<\/p>\n<p>= [9\u00d78!]\/(8!) \u00d7 [20\u00d719\u00d718\u00d717!]\/(17! 3!)<\/p>\n<p>= 9 \u00d7 [20\u00d719\u00d718]\/(3!)<\/p>\n<p>= 9\u00d7 [20\u00d719\u00d718]\/(3\u00d72\u00d71)<\/p>\n<p>= 9 \u00d7 10\u00d719\u00d76<\/p>\n<p>= 10260 ways<\/p>\n<p><strong>(ii)\u00a0<\/strong>a particular student is included.<\/p>\n<p>Number of ways = (choosing 2 professors out of 10 professors) \u00d7 (choosing 2 students out of 19 students)<\/p>\n<p>=\u00a0<sup>10<\/sup>C<sub>2<\/sub>\u00a0\u00d7\u00a0<sup>19<\/sup>C<sub>2<\/sub><\/p>\n<p>By using the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><sup>10<\/sup>C<sub>2<\/sub>\u00a0\u00d7\u00a0<sup>19<\/sup>C<sub>2<\/sub>\u00a0= 10!\/2!(10 \u2013 2)! \u00d7 19!\/2!(19-2)!<\/p>\n<p>= 10!\/(2! 8!) \u00d7 19!\/(2! 17!)<\/p>\n<p>= [10\u00d79\u00d78!]\/(2! 8!) \u00d7 [19\u00d718\u00d717!]\/(17! 2!)<\/p>\n<p>= [10\u00d79]\/2! \u00d7 [19\u00d718]\/(2!)<\/p>\n<p>= [10\u00d79]\/(2\u00d71) \u00d7 [19\u00d718]\/(2\u00d71)<\/p>\n<p>= 5\u00d79 \u00d7 19\u00d79<\/p>\n<p>= 45 \u00d7 171<\/p>\n<p>= 7695 ways<\/p>\n<p><strong>(iii)\u00a0<\/strong>a particular student is excluded.<\/p>\n<p>Number of ways = (choosing 2 professors out of 10 professors) \u00d7 (choosing 3 students out of 19 students)<\/p>\n<p>=\u00a0<sup>10<\/sup>C<sub>2<\/sub>\u00a0\u00d7\u00a0<sup>19<\/sup>C<sub>3<\/sub><\/p>\n<p>By using the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><sup>10<\/sup>C<sub>2<\/sub>\u00a0\u00d7\u00a0<sup>19<\/sup>C<sub>3<\/sub>\u00a0= 10!\/2!(10 \u2013 2)! \u00d7 19!\/3!(19-3)!<\/p>\n<p>= 10!\/(2! 8!) \u00d7 19!\/(3! 16!)<\/p>\n<p>= [10\u00d79\u00d78!]\/(2! 8!) \u00d7 [19\u00d718\u00d717\u00d716!]\/(16! 3!)<\/p>\n<p>= [10\u00d79]\/2! \u00d7 [19\u00d718\u00d717]\/(3!)<\/p>\n<p>= [10\u00d79]\/(2\u00d71) \u00d7 [19\u00d718\u00d717]\/(3\u00d72\u00d71)<\/p>\n<p>= 5\u00d79 \u00d7 19\u00d73\u00d717<\/p>\n<p>= 45 \u00d7 969<\/p>\n<p>= 43605 ways<\/p>\n<p>\u2234\u00a0The required no. of ways are 51300, 10260, 7695, 43605.<\/p>\n<p><strong>6. How many different products can be obtained by multiplying two or more of the numbers 3, 5, 7, 11 (without repetition)?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given that, we need to find the no. of ways of obtaining a product by multiplying two or more from the numbers 3, 5, 7, 11.<\/p>\n<p>Number of ways = (no. of ways of multiplying two numbers) + (no. of ways of multiplying three numbers) + (no. of multiplying four numbers)<\/p>\n<p>=\u00a0<sup>4<\/sup>C<sub>2<\/sub>\u00a0+\u00a0<sup>4<\/sup>C<sub>3<\/sub>\u00a0+\u00a0<sup>4<\/sup>C<sub>4<\/sub><\/p>\n<p>By using 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-5.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 17 \u2013 Combinations image - 5\" \/><\/p>\n<p>= 12\/2 + 4 + 1<\/p>\n<p>= 6 + 4 + 1<\/p>\n<p>= 11<\/p>\n<p>\u2234\u00a0The total number of ways of product is 11 ways.<\/p>\n<p><strong>7. From a class of 12 boys and 10 girls, 10 students are to be chosen for the competition, at least including 4 boys and 4 girls. The 2 girls who won the prizes last year should be included. In how many ways can the selection be made?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>Total number of boys = 12<\/p>\n<p>Total number of girls = 10<\/p>\n<p>Total number of girls for the competition = 10 + 2 = 12<\/p>\n<p>Number of ways = (no. of ways of selecting 6 boys and 2 girls from remaining 12 boys and 8 girls) + (no. of ways of selecting 5 boys and 3 girls from remaining 12 boys and 8 girls) + (no. of ways of selecting 4 boys and 4 girls from remaining 12 boys and 8 girls)<\/p>\n<p>Since two girls have already selected,<\/p>\n<p>= (<sup>12<\/sup>C<sub>6<\/sub>\u00a0\u00d7\u00a0<sup>8<\/sup>C<sub>2<\/sub>) + (<sup>12<\/sup>C<sub>5<\/sub>\u00a0\u00d7\u00a0<sup>8<\/sup>C<sub>3<\/sub>) + (<sup>12<\/sup>C<sub>4<\/sub>\u00a0\u00d7\u00a0<sup>8<\/sup>C<sub>4<\/sub>)<\/p>\n<p>By using 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-6.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 17 \u2013 Combinations image - 6\" \/><\/p>\n<p>= (924 \u00d7 28) + (792 \u00d7 56) + (495 \u00d7 70)<\/p>\n<p>= 25872 + 44352 + 34650<\/p>\n<p>= 104874<\/p>\n<p>\u2234\u00a0The total number of ways of product is 104874 ways.<\/p>\n<p><strong>8.<\/strong>\u00a0<strong>How many different selections of 4 books can be made from 10 different books, if<br \/>(i) There is no restriction<\/strong><\/p>\n<p><strong>(ii) Two particular books are always selected<\/strong><\/p>\n<p><strong>(iii) Two particular books are never selected<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>Total number of books = 10<\/p>\n<p>Total books to be selected = 4<\/p>\n<p><strong>(i)\u00a0<\/strong>there is no restriction<\/p>\n<p>Number of ways = choosing 4 books out of 10 books<\/p>\n<p>=\u00a0<sup>10<\/sup>C<sub>4<\/sub><\/p>\n<p>By using the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><sup>10<\/sup>C<sub>4<\/sub>\u00a0= 10! \/ 4! (10 \u2013 4)!<\/p>\n<p>= 10! \/ (4! 6!)<\/p>\n<p>= [10\u00d79\u00d78\u00d77\u00d76!] \/ (4! 6!)<\/p>\n<p>= [10\u00d79\u00d78\u00d77] \/ (4\u00d73\u00d72\u00d71)<\/p>\n<p>= 10\u00d73\u00d77<\/p>\n<p>= 210 ways<\/p>\n<p><strong>(ii)\u00a0<\/strong>two particular books are always selected<\/p>\n<p>The number of ways = select 2 books out of the remaining 8 books as 2 books are already selected.<\/p>\n<p>=\u00a0<sup>8<\/sup>C<sub>2<\/sub><\/p>\n<p>By using the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><sup>8<\/sup>C<sub>2<\/sub>\u00a0= 8! \/ 2! (8 \u2013 2)!<\/p>\n<p>= 8! \/ (2! 6!)<\/p>\n<p>= [8\u00d77\u00d76!] \/ (2! 6!)<\/p>\n<p>= [8\u00d77] \/ (2\u00d71)<\/p>\n<p>= 4\u00d77<\/p>\n<p>= 28 ways<\/p>\n<p><strong>(iii)\u00a0<\/strong>two particular books are never selected<\/p>\n<p>The number of ways = select 4 books out of the remaining 8 books as 2 books are already removed.<\/p>\n<p>=\u00a0<sup>8<\/sup>C<sub>4<\/sub><\/p>\n<p>By using the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><sup>8<\/sup>C<sub>4<\/sub>\u00a0= 8! \/ 4! (8 \u2013 4)!<\/p>\n<p>= 8! \/ (4! 4!)<\/p>\n<p>= [8\u00d77\u00d76\u00d75\u00d74!] \/ (4! 4!)<\/p>\n<p>= [8\u00d77\u00d76\u00d75] \/ (4\u00d73\u00d72\u00d71)<\/p>\n<p>= 7\u00d72\u00d75<\/p>\n<p>= 70 ways<\/p>\n<p>\u2234\u00a0The required no. of ways are 210, 28, 70.<\/p>\n<p><strong>9. From 4 officers and 8 jawans in how many ways can 6 be chosen (i) to include exactly one officer (ii) to include at least one officer?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>Total number of officers = 4<\/p>\n<p>Total number of jawans = 8<\/p>\n<p>The total number of selections to be made is 6<\/p>\n<p><strong>(i)\u00a0<\/strong>to include exactly one officer<\/p>\n<p>Number of ways = (no. of ways of choosing 1 officer from 4 officers) \u00d7 (no. of ways of choosing 5 jawans from 8 jawans)<\/p>\n<p>= (<sup>4<\/sup>C<sub>1<\/sub>) \u00d7 (<sup>8<\/sup>C<sub>5<\/sub>)<\/p>\n<p>By using 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-7.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 17 \u2013 Combinations image - 7\" \/><\/p>\n<p><strong>(ii)\u00a0<\/strong>to include at least one officer?<\/p>\n<p>Number of ways = (total no. of ways of choosing 6 persons from all 12 persons) \u2013 (no. of ways of choosing 6 persons without any officer)<\/p>\n<p>=\u00a0<sup>12<\/sup>C<sub>6<\/sub>\u00a0\u2013\u00a0<sup>8<\/sup>C<sub>6<\/sub><\/p>\n<p>By using 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-8.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 17 \u2013 Combinations image - 8\" \/><\/p>\n<p>= (11\u00d72\u00d73\u00d72\u00d77) \u2013 (4\u00d77)<\/p>\n<p>= 924 \u2013 28<\/p>\n<p>= 896 ways<\/p>\n<p>\u2234\u00a0The required no. of ways are 224 and 896.<\/p>\n<p><strong>10. A sports team of 11 students is to be constituted, choosing at least 5 from class XI and at least 5 from class XII. If there are 20 students in each of these classes, in how many ways can the teams be constituted?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>Total number of students in XI = 20<\/p>\n<p>Total number of students in XII = 20<\/p>\n<p>Total number of students to be selected in a team = 11 (with at least 5 from class XI and 5 from class XII)<\/p>\n<p>Number of ways = (No. of ways of selecting 6 students from class XI and 5 students from class XII) + (No. of ways of selecting 5 students from class XI and 6 students from class XII)<\/p>\n<p>= (<sup>20<\/sup>C<sub>6<\/sub>\u00a0\u00d7\u00a0<sup>20<\/sup>C<sub>5<\/sub>) + (<sup>20<\/sup>C<sub>5<\/sub>\u00a0\u00d7\u00a0<sup>20<\/sup>C<sub>6<\/sub>)<\/p>\n<p>= 2 (<sup>20<\/sup>C<sub>6<\/sub>\u00a0\u00d7\u00a0<sup>20<\/sup>C<sub>5<\/sub>) ways<\/p>\n<p><strong>11. A student has to answer 10 questions, choosing at least 4 from each of part A and part B. If there are 6 questions in part A and 7 in part B, in how many ways can the student choose 10 questions?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>Total number of questions = 10<\/p>\n<p>Questions in part A = 6<\/p>\n<p>Questions in part B = 7<\/p>\n<p>Number of ways = (No. of ways of answering 4 questions from part A and 6 from part B) + (No. of ways of answering 5 questions from part A and 5 questions from part B) + (No. of ways of answering 6 questions from part A and 4 from part B)<\/p>\n<p>= (<sup>6<\/sup>C<sub>4<\/sub>\u00a0\u00d7\u00a0<sup>7<\/sup>C<sub>6<\/sub>) + (<sup>6<\/sup>C<sub>5<\/sub>\u00a0\u00d7\u00a0<sup>7<\/sup>C<sub>5<\/sub>) + (<sup>6<\/sup>C<sub>6<\/sub>\u00a0\u00d7\u00a0<sup>7<\/sup>C<sub>4<\/sub>)<\/p>\n<p>By using 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-9.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 17 \u2013 Combinations image - 9\" \/><\/p>\n<p>= (15\u00d77) + (6\u00d721) + (1\u00d735)<\/p>\n<p>= 105 + 126 + 35<\/p>\n<p>= 266<\/p>\n<p>\u2234\u00a0The total no. of ways of answering 10 questions is 266 ways.<\/p>\n<p><strong>12. In an examination, a student to answer 4 questions out of 5 questions; questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make a choice.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>Total number of questions = 5<\/p>\n<p>Total number of questions to be answered = 4<\/p>\n<p>The number of ways = we need to answer 2 questions out of the remaining 3 questions as 1 and 2 are compulsory.<\/p>\n<p>=\u00a0<sup>3<\/sup>C<sub>2<\/sub><\/p>\n<p>By using the formula,<\/p>\n<p><sup>n<\/sup>C<sub>r<\/sub>\u00a0= n!\/r!(n \u2013 r)!<\/p>\n<p><sup>3<\/sup>C<sub>2<\/sub>\u00a0= 3!\/2!(3 -2)!<\/p>\n<p>= 3! \/ (2! 1!)<\/p>\n<p>= [3\u00d72\u00d71] \/ (2\u00d71)<\/p>\n<p>= 3<\/p>\n<p>\u2234 The no. of ways to answer the questions is 3.<\/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 17.2. If you have any queries feel free to ask in the comment section.\u00a0<\/p>\n<div class=\"kopyk-content\">\n<h3><span class=\"ez-toc-section\" id=\"faq-rd-sharma-solutions-class-11-maths-chapter-17-exercise-172\"><\/span>FAQ: RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.2<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<\/div>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630413897363\" 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 id=\"faq-question-1630413924873\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-download-rd-sharma-solutions-exercise-172-class-11-maths-chapter-17-pdf-for-free\"><\/span>Can I download RD Sharma Solutions Exercise 17.2 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 17.2 PDF free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630413938693\" 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-172-pdf-on-my-smartphone\"><\/span>Can I open RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.2 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 17.2 PDF on any device.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630413942205\" 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>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.2: We previously studied the arrangement of a fixed number of objects, while acquiring some or all of them. Part or all of the elements, regardless of their arrangement, each different option is called a combination. Students who wish to learn by themselves and solve problems &#8230; <a title=\"RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.2 (Updated 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-17-exercise-17-2\/\" aria-label=\"More on RD Sharma Solutions Class 11 Maths Chapter 17 Exercise 17.2 (Updated 2024)\">Read more<\/a><\/p>\n","protected":false},"author":244,"featured_media":121921,"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\/67983"}],"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=67983"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67983\/revisions"}],"predecessor-version":[{"id":504315,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67983\/revisions\/504315"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/121921"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=67983"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=67983"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=67983"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}