{"id":6025,"date":"2025-06-09T14:17:38","date_gmt":"2025-06-09T14:17:38","guid":{"rendered":"https:\/\/thecompanyboy.com\/?p=6025"},"modified":"2025-06-09T14:17:38","modified_gmt":"2025-06-09T14:17:38","slug":"rs-aggarwal-quantitative-aptitude-pdf-download-permutations-combinations-and","status":"publish","type":"post","link":"https:\/\/www.reilsolar.com\/drive\/rs-aggarwal-quantitative-aptitude-pdf-download-permutations-combinations-and\/","title":{"rendered":"RS Aggarwal Quantitative Aptitude PDF Free Download: PERMUTATIONS AND COMBINATIONS"},"content":{"rendered":"

PERMUTATIONS AND COMBINATIONS<\/h1>\n

IMPORTANT FACTS AND FORMULAE<\/h2>\n

Factorial Notation:<\/strong> Let n be a positive integer.\u00a0 Then, factorial n, denoted by n!\u00a0 is defined as:<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 n! = n(n-1)(n-2)……..3.2.1.<\/strong><\/p>\n

\u00a0<\/strong>Examples:<\/strong> (i) 5! = (5x 4 x 3 x 2 x 1) = 120; (ii) 4! = (4x3x2x1) = 24 etc.<\/p>\n

We define, 0! = 1.<\/p>\n

Permutations:<\/strong>\u00a0 The different arrangements of a given number of things by taking some or all at a time, are called permutations.<\/p>\n

Ex. 1.<\/strong>All permutations (or arrangements) made with the letters a, b, c by taking\u00a0 two at a time are: (ab, ba, ac, bc, cb).<\/strong><\/p>\n

Ex. 2.<\/strong>All permutations made with the letters a,b,c, taking all at a time are:<\/p>\n

(abc, acb, bca, cab, cba).<\/strong><\/p>\n

Number of Permutations: <\/strong>Number of all permutations of n things, taken r\u00a0 at a time, given by:<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 n<\/sup>Pr\u00a0 <\/sub>= n(n-1)(n-2)…..(n-r+1) = n!\/(n-r)!<\/strong><\/p>\n

Examples:<\/strong> (i) 6<\/sup>p2<\/sub> = (6×5) = 30. (ii) 7<\/sup>p3 <\/sub>= (7x6x5) = 210.<\/p>\n

Cor. Number of all permutations of n things, taken all at a time = n!<\/strong><\/p>\n

\u00a0<\/strong>An Important Result: <\/strong>If there are n objects of\u00a0 which p1<\/sub> are alike of\u00a0 one kind; p2<\/sub> are alike of another kind; p3<\/sub> are alike of third kind and\u00a0 so on and p\u00adr <\/sub>are alike of rth kind, such that (p1<\/sub>+p2<\/sub>+…….pr<\/sub>) = n.<\/p>\n

Then, number of permutations of these n objects is:<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 n!\u00a0 \/\u00a0 (p1<\/sub>!).p2<\/sub>!)……(pr<\/sub>!)<\/strong><\/p>\n

Combinations:<\/strong> Each of the different groups or selections which can be formed by taking some or all of a number of objects, is called a combination.<\/p>\n

Ex. 1.<\/strong> Suppose we want to select two out of three boys A, B, C.\u00a0 Then, possible selections are AB, BC and CA.<\/p>\n

Note that AB and BA represent the same selection.<\/p>\n

Ex. 2.<\/strong> All the combinations formed by a, b, c, taking two at a time are ab, bc, ca.<\/strong><\/p>\n

Ex. 3.<\/strong> The only combination that can be formed of three letters a, b, c taken all at a time is abc.<\/strong><\/p>\n

Ex. 4<\/strong>. Various groups of 2 out of four presons A, B, C, D are:<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 AB, AC, AD, BC, BD, CD.<\/strong><\/p>\n

Ex. 5.<\/strong> Note that ab and ba are two different permutations but they represent\u00a0 the same combination.<\/p>\n

Number of Combinations:<\/strong> The number of all combination of n things,<\/p>\n

taken r at a time is:<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 n<\/sup>Cr <\/sub>= n! \/ (r!)(n-r)! = n(n-1)(n-2)…..to r factors \/ r!<\/strong><\/p>\n

Note that: n<\/sup>cr <\/sub>\u00ad = 1 and n<\/sup>c0<\/sub> = 1.<\/p>\n

An Important Result:\u00a0 n<\/sup>cr<\/sub> = n<\/sup>c(n-r)<\/sub>.<\/p>\n

Example:<\/strong>\u00a0 (i)\u00a0 11<\/sup>c4<\/sub> = (11x10x9x8)\/(4x3x2x1) = 330.<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (ii) 16<\/sup>c13<\/sub> = 16<\/sup>c(16-13) <\/sub>= 16x15x14\/3! = 16x15x14\/3x2x1 = 560.<\/p>\n

SOLVED EXAMPLES<\/h2>\n

Ex. 1.<\/strong> Evaluate: 30!\/28!<\/p>\n

Sol.<\/strong>\u00a0 We have, 30!\/28! = 30x29x(28!)\/28! = (30×29) = 870.<\/p>\n

Ex. 2<\/strong>. Find the value of (i) 60<\/sup>p3<\/sub>\u00a0 (ii) 4<\/sup>p4<\/sub><\/p>\n

Sol.\u00a0 <\/strong>(i) 60<\/sup>p3<\/sub> = 60!\/(60-3)! = 60!\/57! = 60x59x58x(57!)\/57! = (60x59x58) = 205320.<\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 (ii) 4<\/sup>p4<\/sub> = 4! = (4x3x2x1) = 24.<\/p>\n

Ex. 3.<\/strong> Find the vale of (i) 10<\/sup>c3 <\/sub>(ii) 100<\/sup>c98<\/sub> (iii) 50<\/sup>c50<\/sub><\/p>\n

Sol.<\/strong> (i) 10<\/sup>c3 <\/sub>= 10x9x8\/3! = 120.<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (ii) 100<\/sup>c98<\/sub> = 100<\/sup>c(100-98)<\/sub> = 100×99\/2! = 4950.<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (iii) 50<\/sup>c50<\/sub> = 1.\u00a0\u00a0\u00a0\u00a0\u00a0 [n<\/sup>cn<\/sub> = 1]<\/p>\n

Ex. 4.<\/strong> How many words can be formed by using all letters of the word \u201cBIHAR\u201d<\/p>\n

Sol.<\/strong>\u00a0 The word BIHAR contains 5 different letters.<\/p>\n

Required number of words = 5<\/sup>p5<\/sub> = 5! = (5x4x3x2x1) = 120.<\/p>\n

Ex. 5.<\/strong> How many words can be formed by using all letters of the word \u2018DAUGHTER\u2019 so that the vowels always come together?<\/p>\n

Sol.<\/strong>\u00a0 Given word contains 8 different letters.\u00a0 When the vowels AUE are always together, we may suppose them to form an entity, treated as one letter.<\/p>\n

Then, the letters to be arranged are DGNTR (AUE).<\/p>\n

Then 6 letters to be arranged in 6<\/sup>p6<\/sub> = 6! = 720 ways.<\/p>\n

The vowels in the group (AUE) may be arranged in 3! = 6 ways.<\/p>\n

Required number of words = (720×6) = 4320.<\/p>\n

Ex. 6.<\/strong> How many words can be formed from the letters of the word \u2018EXTRA\u2019 so that the vowels are never together?<\/p>\n

Sol.<\/strong>\u00a0 The given word contains 5 different letters.<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Taking the vowels EA together, we treat them as one letter.<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Then, the letters to be arranged are XTR (EA).<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 These letters can be arranged in 4! = 24 ways.<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The vowels EA may be arranged amongst themselves in 2! = 2 ways.<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Number of words, each having vowels together = (24×2) = 48 ways.<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Total number of words formed by using all the letters of the given words<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 5! = (5x4x3x2x1) = 120.<\/p>\n

Number of words, each having vowels never together = (120-48) = 72.<\/p>\n

Ex. 7.<\/strong>\u00a0 How many words can be formed from the letters of the word \u2018DIRECTOR\u2019<\/p>\n

So that the vowels are always together?<\/p>\n

Sol.<\/strong> In the given word, we treat the vowels IEO as one letter.<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Thus, we have DRCTR (IEO).<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 This group has 6 letters of which R occurs 2 times and others are different.<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Number of ways of arranging these letters = 6!\/2! = 360.<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Now 3 vowels can be arranged among themselves in 3! = 6 ways.<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Required number of ways = (360×6) = 2160.<\/p>\n

Ex. 8.<\/strong>\u00a0 In how many ways can a cricket eleven be chosen out of a batch of<\/p>\n

15 players ?<\/p>\n

Sol.<\/strong>\u00a0 Required number of ways = 15<\/sup>c11 <\/sub>= 15<\/sup>c(15-11)<\/sub> = 11<\/sup>c4<\/sub><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 15x14x13x12\/4x3x2x1 = 1365.<\/p>\n

Ex. 9<\/strong>.\u00a0 In how many ways, a committee of 5 members can be selected from<\/p>\n

6 men and 5 ladies, consisting of 3 men and 2 ladies?<\/p>\n

Sol.<\/strong>\u00a0 (3 men out 6) and (2 ladies out of 5) are to be chosen.<\/p>\n

Required number of ways = (6<\/sup>c3<\/sub>x5<\/sup>c2<\/sub>) = [6x5x4\/3x2x1] x [5×4\/2×1] = 200.\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n

Accessing copyrighted materials like R.S. Aggarwal’s “Quantitative Aptitude” without proper authorization is not advisable, as it may violate intellectual property rights. However, to assist you in studying Permutations and Combinations<\/strong>, here are some reputable resources:<\/p>\n

    \n
  1. \n

    Vedantu<\/strong>: Offers free solutions and explanations for Class 11 Maths Chapter 8 on Permutations, based on R.S. Aggarwal’s textbook. These solutions can help you understand the concepts and practice problems effectively.<\/p>\n

    Clear IIT Medical<\/strong>: Provides free PDF downloads of solutions for R.S. Aggarwal’s Class 11 Maths Chapter 8 on Permutations. This resource includes detailed explanations and solved examples to aid your learning.<\/li>\n<\/ol>\n

    For comprehensive learning, consider purchasing the complete book from authorized retailers or accessing it through legitimate educational platforms. This ensures you have access to accurate and complete information while respecting intellectual property rights.<\/p>\n

    RS Aggarwal Quantitative Aptitude PDF Free Download: PERMUTATIONS AND COMBINATIONS<\/a><\/h3>\n

    FOR COMPETITIVE EXAMINATIONS FULLY SOLVED<\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

    PERMUTATIONS AND COMBINATIONS IMPORTANT FACTS AND FORMULAE Factorial Notation: Let n be a positive integer.\u00a0 Then, factorial n, denoted by n!\u00a0 is defined as: \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 n! = n(n-1)(n-2)……..3.2.1. \u00a0Examples: (i) 5! = (5x 4 x 3 x 2 x 1) = 120; (ii) 4! = (4x3x2x1) = 24 etc. We define, 0! = 1. Permutations:\u00a0 […]<\/p>\n","protected":false},"author":41,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[126,127],"tags":[],"class_list":["post-6025","post","type-post","status-publish","format-standard","hentry","category-rs-aggarwal-quantitative-aptitude","category-rs-aggarwal-quantitative-aptitude-pdf"],"_links":{"self":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/6025","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/users\/41"}],"replies":[{"embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/comments?post=6025"}],"version-history":[{"count":0,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/6025\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/media?parent=6025"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/categories?post=6025"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/tags?post=6025"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}