{"id":23399,"date":"2022-09-02T09:13:05","date_gmt":"2022-09-02T03:43:05","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=23399"},"modified":"2022-09-02T09:13:05","modified_gmt":"2022-09-02T03:43:05","slug":"inter-2nd-year-maths-2a-permutations-and-combinations-solutions-ex-5e","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/inter-2nd-year-maths-2a-permutations-and-combinations-solutions-ex-5e\/","title":{"rendered":"Inter 2nd Year Maths 2A Permutations and Combinations Solutions Ex 5(e)"},"content":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions<\/a> Inter 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(e) will help students to clear their doubts quickly.<\/p>\n

Intermediate 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(e)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nIf n<\/sup>C4<\/sub> = 210, find n.
\nSolution:
\nn<\/sup>Cr<\/sub> = \\(\\frac{n !}{(n-r) ! r !}\\) = \\(\\frac{n(n-1)(n-2) \\ldots \\ldots[n-(r-1)]}{1.2 .3 \\ldots \\ldots \\ldots . . . r}\\)
\nSolution:
\nn<\/sup>C4<\/sub> = 210
\n\u21d2 \\(\\frac{n(n-1)(n-2)(n-3)}{1.2 .3 .4}=10 \\times 21^n C_{2 r-1}\\)
\n\u21d2 n(n – 1) (n – 2) (n – 3) = 10 \u00d7 21 \u00d7 1 \u00d7 2 \u00d7 3 \u00d7 4
\n\u21d2 n(n – 1) (n – 2) (n – 3) = 10 \u00d7 7 \u00d7 3 \u00d7 2 \u00d7 3 \u00d7 4
\n\u21d2 n(n – 1) (n – 2) (n – 3) = 10 \u00d7 9 \u00d7 8 \u00d7 7
\n\u21d2 n = 10<\/p>\n

Question 2.
\nIf 12<\/sup>Cr<\/sub> = 495, find the possible values of r.
\nSolution:
\nHint: n<\/sup>Cr<\/sub> = n<\/sup>Cn-r<\/sub>
\n12<\/sup>Cr<\/sub> = 495
\n= 5 \u00d7 99
\n= 11 \u00d7 9 \u00d7 5
\n= \\(\\frac{12 \\times 11 \\times 9 \\times 5 \\times 2}{12 \\times 2}\\)
\n= \\(\\frac{12 \\times 11 \\times 10 \\times 9}{1.2 .3 .4}\\)
\n= 12<\/sup>C4<\/sub> or 12<\/sup>C8<\/sub>
\n\u2234 r = 4 or 8<\/p>\n

\"Inter<\/p>\n

Question 3.
\nIf 10 . n<\/sup>C2<\/sub> = 3 . n+1<\/sup>C3<\/sub>, find n.
\nSolution:
\n10 . n<\/sup>C2<\/sub> = 3 . n+1<\/sup>C3<\/sub>
\n\u21d2 10 \u00d7 \\(\\frac{n(n-1)}{1.2}=\\frac{3(n+1)(n-1)}{1.2 .3}\\)
\n\u21d2 10 = n + 1
\n\u21d2 n = 9<\/p>\n

Question 4.
\nIf n<\/sup>Pr<\/sub> = 5040 and n<\/sup>Cr<\/sub> = 210, find n and r.
\nSolution:
\nHint: n<\/sup>Pr<\/sub> = r! n<\/sup>Cr<\/sub> and n<\/sup>Pr<\/sub> = n(n – 1) (n – 2) ……. [n – (r – 1)]
\nn<\/sup>Pr<\/sub> = 5040, n<\/sup>Cr<\/sub> = 210
\nr! = \\(\\frac{{ }^n P_r}{{ }^n C_r}=\\frac{5040}{210}=\\frac{504}{21}\\) = 24 = 4!
\n\u2234 r = 4
\nn<\/sup>Pr<\/sub> = 5040
\nn<\/sup>P4<\/sub> = 5040
\n= 10 \u00d7 504
\n= 10 \u00d7 9 \u00d7 56
\n= 10 \u00d7 9 \u00d7 8 \u00d7 7
\n= 10<\/sup>P4<\/sub>
\n\u2234 n = 10
\n\u2234 n = 10, r = 4<\/p>\n

Question 5.
\nIf n<\/sup>C4<\/sub> = n<\/sup>C6<\/sub>, find n.
\nSolution:
\nn<\/sup>Cr<\/sub> = n<\/sup>Cs<\/sub> \u21d2 r = s or r + s = n
\nn<\/sup>C4<\/sub> = n<\/sup>C6<\/sub>
\n\u2234 n = 4 + 6 = 10, (\u2235 4 \u2260 6)<\/p>\n

Question 6.
\nIf 15<\/sup>C2r-1<\/sub> = 15<\/sup>C2r+4<\/sub>, find r.
\nSolution:
\n15<\/sup>C2r-1<\/sub> = 15<\/sup>C2r+4<\/sub>
\n2r – 1 = 2r + 4 or (2r – 1) + (2r + 4) = 15
\n(2r – 1) + (2r + 4) = 15
\n\u21d2 4r + 3 = 15
\n\u21d2 4r = 12
\n\u21d2 r = 3
\n\u2234 2r – 1 = 2r + 4
\n\u21d2 -1 = 4 which is impossible
\n\u2234 r = 3<\/p>\n

\"Inter<\/p>\n

Question 7.
\nIf 17<\/sup>C2t+1<\/sub> = 17<\/sup>C3t-5<\/sub>, find t.
\nSolution:
\n17<\/sup>C2t+1<\/sub> = 17<\/sup>C3t-5<\/sub>
\n2t + 1 = 3t – 5 or (2t + 1) + (3t – 5) = 17
\n\u21d2 1 + 5 = t or 5t = 21
\n\u21d2 t = 6 or t = \\(\\frac{21}{5}\\) which is not an integer
\n\u2234 t = 6<\/p>\n

Question 8.
\nIf 12<\/sup>Cr+1<\/sub> = 12<\/sup>C3r-5<\/sub>, find r.
\nSolution:
\n12<\/sup>Cr+1<\/sub> = 12<\/sup>C3r-5<\/sub>
\n\u21d2 r + 1 = 3r – 5 or (r + 1) + (3r – 5) = 12
\n\u21d2 1 + 5 = 2r or 4r – 4 = 12
\n\u21d2 2r = 6 or 4r = 16
\n\u21d2 r = 3 or r = 4
\n\u2234 r = 3 or 4<\/p>\n

Question 9.
\nIf 9<\/sup>C3<\/sub> + 9<\/sup>C5<\/sub> = 10<\/sup>Cr<\/sub> then find r.
\nSolution:
\nn<\/sup>Cr<\/sub> = n<\/sup>Cn-r<\/sub>
\n10<\/sup>Cr<\/sub> = 9<\/sup>C3<\/sub> + 9<\/sup>C5<\/sub>
\n\u2234 9<\/sup>C3<\/sub> + 9<\/sup>C5<\/sub> = 9<\/sup>C3<\/sub> + 9<\/sup>C5<\/sub> = 10<\/sup>C6<\/sub> or 10<\/sup>C4<\/sub> = 10<\/sup>Cr<\/sub> (given)
\n\u21d2 r = 4 or 6<\/p>\n

Question 10.
\nFind the number of ways of forming a committee of 5 members from 6 men and 3 ladies.
\nSolution:
\nTotal number of persons = 6 + 3 = 9
\n\u2234 Number of ways of forming a committee of 5 members from 6 men and 3 ladies = 9<\/sup>C5<\/sub>
\n= \\(\\frac{9 \\times 8 \\times 7 \\times 6 \\times 5}{5 \\times 4 \\times 3 \\times 2 \\times 1}\\)
\n= 126<\/p>\n

\"Inter<\/p>\n

Question 11.
\nIn question no. 10, how many committees contain atleast two ladies?
\nSolution:
\nSince a committee contains atleast 2 ladies, the members of the committee may be of the following two types.
\n(i) 3 men, 2 ladies
\n(ii) 2 men, 3 ladies
\nThe number of selections in the first type = 6<\/sup>C3<\/sub> \u00d7 3<\/sup>C2<\/sub>
\n= 20 \u00d7 3
\n= 60
\nThe number of selections in the second type = 6<\/sup>C2<\/sub> \u00d7 3<\/sup>C3<\/sub>
\n= 15 \u00d7 1
\n= 15
\n\u2234 The required number of ways of selecting the committee containing atleast 2 ladies = 60 + 15 = 75.<\/p>\n

Question 12.
\nIf n<\/sup>C5<\/sub> = n<\/sup>C6<\/sub>, then find 13<\/sup>Cn<\/sub>.
\nSolution:
\n\u2235 n<\/sup>C5<\/sub> = n<\/sup>C6<\/sub>
\n\u21d2 n = 6 + 5 = 11
\n13<\/sup>Cn<\/sub> = 13<\/sup>C11<\/sub> = 13<\/sup>C2<\/sub>
\n= \\(\\frac{13 \\times 12}{1 \\times 2}\\)
\n= 78<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nProve that for 3 \u2264 r \u2264 n, (n-3)<\/sup>Cr<\/sub>\u00a0. (n-3)<\/sup>C(r-1)<\/sub> + 3 . (n-3)<\/sup>C(r-2)<\/sub> + 3 . (n-3)<\/sup>C(r-3)<\/sub> = n<\/sup>Cr<\/sub>
\nSolution:
\n\"Inter<\/p>\n

Question 2.
\nFind the value of 10<\/sup>C5<\/sub> + 2 . 10<\/sup>C4<\/sub> + 10<\/sup>C3<\/sub>
\nSolution:
\nHint: n<\/sup>Cr<\/sub> + n<\/sup>Cr-1<\/sub> = (n+1)<\/sup>Cr<\/sub>
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 3.
\nSimplify 34<\/sup>C5<\/sub> + \\(\\sum_{r=0}^4{ }^{(38-r)} C_4\\)
\nSolution:
\n\"Inter<\/p>\n

Question 4.
\nIn a class, there are 30 students. If each student plays a chess game with each of the other students then find the total number of chess games played by them.
\nSolution:
\nNumber of students in a class = 30
\nSince each student plays a chess game with each of the other students, the total number of chess games played by them = 30<\/sup>C2<\/sub> = 435<\/p>\n

Question 5.
\nFind the number of ways of selecting 3 girls and 3 boys out of 7 girls and 6 boys.
\nSolution:
\nThe number of ways of selecting 3 girls and 3 boys Out of 7 girls and 6 boys = 7<\/sup>C3<\/sub> \u00d7 6<\/sup>C3<\/sub>
\n= 35 \u00d7 20
\n= 700<\/p>\n

Question 6.
\nFind the number of ways of selecting a committee of 6 members out of 10 members always including a specified member.
\nSolution:
\nSince a specified member is always included in a committee, the remaining 5 members can be selected from the remaining 9 members in 9<\/sup>C5<\/sub> ways.
\n\u2234 Required number of ways selecting a committee = 9<\/sup>C5<\/sub> = 126<\/p>\n

Question 7.
\nFind the number of ways of selecting 5 books from 9 different mathematics books such that a particular book is not included.
\nSolution:
\nSince a particular book is not included in the selection, the 5 books can be selected from the remaining 8 books in 8<\/sup>C5<\/sub> ways.
\n\u2234 The required number of ways of selecting 5 books = 8<\/sup>C5<\/sub> = 56<\/p>\n

\"Inter<\/p>\n

Question 8.
\nFind the number of ways of selecting 3 vowels and 2 consonants from the letters of the word EQUATION.
\nSolution:
\nThe word EQUATION contains 5 vowels and 3 consonants.
\nThe 3 vowels can be selected from 5 vowels in 5<\/sup>C3<\/sub> = 10 ways.
\nThe 2 consonants can be selected from 3 consonants in 3<\/sup>C2<\/sub> = 3 ways.
\n\u2234 The required number of ways of selecting 3 vowels and 2 consonants = 10 \u00d7 3 = 30<\/p>\n

Question 9.
\nFind the number of diagonals of a polygon with 12 sides.
\nSolution:
\nThe number of diagonals of a polygon with sides = \\(\\frac{n(n-3)}{2}\\)
\n= \\(\\frac{12(12-3)}{2}\\)
\n= 54<\/p>\n

Question 10.
\nIf n persons are sitting in a row, find the number of ways of selecting two persons, who are sitting adjacent to each other.
\nSolution:
\nThe number of ways of selecting 2 persons out of n persons sitting in a row, who are sitting adjacent to each other = n – 1<\/p>\n

\"Inter<\/p>\n

Question 11.
\nFind the number of ways of giving away 4 similar coins to 5 boys if each boy can be given any number (less than or equal to 4) of coins.
\nSolution:
\nThe 4 similar coins can be divided into different groups as follows.
\n(i) One group containing 4 coins
\n(ii) Two groups containing 1, 3 coins respectively
\n(iii) Two groups containing 2, 2 coins respectively
\n(iv) Two groups containing 3, 1 coins respectively
\n(v) Three groups containing 1, 1, 2 coins respectively
\n(vi) Three groups containing 1, 2, 1 coins respectively
\n(vii) Three groups containing 2, 1, 1 coins respectively
\n(viii) Four groups containing 1, 1, 1, 1 coins respectively
\nthese groups can given away to 5 boys in = \\({ }^5 C_1+2 \\times{ }^5 C_2+{ }^5 C_2+{ }^5 C_3 \\times \\frac{3 !}{2 !}+{ }^5 C_4\\)
\n= 5 + 20 + 10 + 30 + 5
\n= 70 ways<\/p>\n

III.<\/span><\/p>\n

Question 1.
\nProve that \\(\\frac{{ }^{4 n} C_{2 n}}{{ }^{2 n} C_n}=\\frac{1.3 .5 \\ldots \\ldots(4 n-1)}{\\{1.3 .5 \\ldots \\ldots(2 n-1)\\}^2}\\)
\nSolution:
\n\"Inter<\/p>\n

Question 2.
\nIf a set A has 12 elements, find the number of subsets of A having
\n(i) 4 elements
\n(ii) Atleast 3 elements
\n(iii) Atmost 3 elements
\nSolution:
\nNumber of elements in set A = 12
\n(i) Number of subsets of A with exactly 4 elements = 12<\/sup>C4<\/sub> = 495<\/p>\n

(ii) The required subset contains atleast 3 elements.
\nThe number of subsets of A with exactly 0 elements is 12<\/sup>C0<\/sub>
\nThe number of subsets of A with exactly 1 element is 12<\/sup>C1<\/sub>
\nThe number of subsets of A with exactly 2 elements is 12<\/sup>C2<\/sub>
\nTotal number of subsets of A formed = 212<\/sup>
\n\u2234 Number of subsets of A with atleast 3 elements = (Total number of subsets) – (number of subsets contains 0 or 1 or 2 elements)
\n= 212<\/sup> – (12<\/sup>C0<\/sub> + 12<\/sup>C1<\/sub> + 12<\/sup>C2<\/sub>)
\n= 4096 – (1 + 12 + 66)
\n= 4096 – 79
\n= 4017<\/p>\n

(iii) The required subset contains atmost 3 elements
\ni.e., it may contain 0 or 1 or 2 or 3 elements.
\nThe number of subsets of A with exactly 0 elements is 12<\/sup>C0<\/sub>
\nThe number of subsets of A with exactly 1 element is 12<\/sup>C1<\/sub>
\nThe number of subsets of A with exactly 2 elements is 12<\/sup>C2<\/sub>
\nThe number of subsets of A with exactly 3 elements is 12<\/sup>C3<\/sub>
\n\u2234 Number of subsets of A with atmost 3 elements = 12<\/sup>C0<\/sub> + 12<\/sup>C1<\/sub> + 12<\/sup>C2<\/sub> + 12<\/sup>C3<\/sub>
\n= 1 + 12 + 66 + 220
\n= 299<\/p>\n

\"Inter<\/p>\n

Question 3.
\nFind the number of ways of selecting a cricket team of 11 players from 7 batsmen and 6 bowlers such that there will be atleast 5 bowlers in the team.
\nSolution:
\nSince the team consists of at least 5 bowlers, the selection may be of the following types.
\n\"Inter
\nThe number of selections in the first type = 7<\/sup>C6<\/sub> \u00d7 6<\/sup>C5<\/sub>
\n= 7 \u00d7 6
\n= 42
\nThe number of selections in the second type = 7<\/sup>C5<\/sub> \u00d7 6<\/sup>C6<\/sub>
\n= 21 \u00d7 1
\n= 21
\n\u2234 The required number of ways selecting the cricket team = 42 + 21 = 63<\/p>\n

Question 4.
\nIf 5 vowels and 6 consonants are given, then how many 6-letter words can be formed with 3 vowels and 3 consonants?
\nSolution:
\nNo. of vowels given = 5
\nNo.of consonants given = 6
\nWe have to form a 6-letter word with 3 vowels and 3 consonants from given letters.
\n3 vowels can select from 5 in 5<\/sup>C3<\/sub> ways.
\n3 consonants can select from 6 in 6<\/sup>C3<\/sub> ways.
\nTotal No. of words = 5<\/sup>C3<\/sub> \u00d7 6<\/sup>C3<\/sub> \u00d7 6! = 1,44,000<\/p>\n

Question 5.
\nThere are 8 railway stations along a railway line. In how many ways can a train be stopped at 3 of these stations such that no two of them are consecutive?
\nSolution:
\nNumber of ways of selecting 3 stations out of 8 = 8<\/sup>C3<\/sub> = 56
\nNumber of ways of selecting 3 out of 8 stations such that 3 are consecutive = 6
\nNumber of ways of selecting 3 out of 8 stations such that 2 of them are consecutive = 2 \u00d7 5 + 5 \u00d7 4
\n= 10 + 20
\n= 30
\n\u2234 Number of ways for a train to be stopped at 3 of 8 stations such that no two of them are consecutive = 56 – (6 + 30) = 20<\/p>\n

\"Inter<\/p>\n

Question 6.
\nFind the number of ways of forming a committee of 5 members out of 6 Indians and 5 Americans so that always the Indians will be in majority in the committee.
\nSolution:
\nSince the committee contains the majority of Indians, the members of the committee may be of the following types.
\n\"Inter
\nThe number of selections in type I = 6<\/sup>C5<\/sub> \u00d7 5<\/sup>C0<\/sub> = 6 \u00d7 1 = 6
\nThe number of selections in type II = 6<\/sup>C4<\/sub> \u00d7 5<\/sup>C1<\/sub> = 15 \u00d7 5 = 75
\nThe number of selections in type III = 6<\/sup>C3<\/sub> \u00d7 5<\/sup>C2<\/sub> = 20 \u00d7 10 = 200
\n\u2234 The required number ways of selecting a committee = 6 + 75 + 200 = 281.<\/p>\n

Question 7.
\nA question paper is divided into 3 sections A, B, C Containing 3, 4, 5 questions respectively. Find the number of ways of attempting 6 questions choosing at least one from each section.
\nSolution:
\nFirst Method: The selection of a question may be of the following
\n\"Inter
\nTotal No. of ways of attempting 6 questions
\n\"Inter
\nSecond Method:
\nRequired No.of attempting 6 questions = Total no. of arrangements – selection except question from C – selection except Q from A – selection except Q from B
\n= 12<\/sup>C6<\/sub> – 7<\/sup>C6<\/sub> – 9<\/sup>C6<\/sub> – 6<\/sup>C6<\/sub>
\n= 805<\/p>\n

Question 8.
\nFind the number of ways in which 12 things be
\n(i) divided into 4 equal groups
\n(ii) distributed to 4 persons equally.
\nSolution:
\n(i) The number of ways in which 12 things be divided into 4 equal groups = \\(\\frac{12 !}{3 ! 3 ! 3 ! 3 ! 4 !}\\) = \\(\\frac{12 !}{(3 !)^4 4 !}\\)
\n(ii) The number of ways in which 12 things be distributed to 4 persons equally = \\(\\frac{12 !}{3 ! 3 ! 3 ! 3 !}\\) = \\(\\frac{12 !}{(3 !)^4}\\)<\/p>\n

\"Inter<\/p>\n

Question 9.
\nA class contains 4 boys and g girls. Every Sunday, five students with atleast 3boys go for a picnic. A different group is being sent every week. During the picnic, the class teacher gives each girl in the group a doll. If the total number of dolls distributed is 85, find g.
\nSolution:
\nNo. of boys = 4
\nNo. of girls = g
\nSince there should be atleast 3 boys it can be done in 2 ways as shown in the table
\n\"Inter
\nThe number of girls in G1<\/sub> = [4<\/sup>C3<\/sub> \u00d7 g<\/sup>C2<\/sub>] \u00d7 2
\nSince each group contains 2 girls
\nThe number of girls in G2<\/sub> = [4<\/sup>C3<\/sub> \u00d7 g<\/sup>C2<\/sub>] \u00d7 1
\nSince each group contains 1 girl.
\nGiven no. of dolls distributed = 85
\n\u21d2 [4<\/sup>C3<\/sub> \u00d7 g<\/sup>C2<\/sub>] \u00d7 2 + [4<\/sup>C4<\/sub> \u00d7 g<\/sup>C1<\/sub>] \u00d7 1 = 85
\n\u21d2 4 . \\(\\frac{g(g-1)}{2}\\) \u00d7 2 + 1 . g . 1 = 85
\n\u21d2 4g2<\/sup>\u00a0– 4g + g – 85 = 0
\n\u21d2 4g2<\/sup> – 3g – 85 = 0
\n\u21d2 4g2<\/sup> – 20g + 17g – 85 = 0
\n\u21d2 4g(g – 5) + 17(g – 5) = 0
\n\u21d2 (g – 5)(4g + 17) = 0
\nSince g \u2260 \\(\\frac{-17}{4}\\)
\n\u2234 g = 5
\nHence No. of girls = 5<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(e) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(e) I. Question 1. If nC4 = 210, find n. Solution: nCr = = Solution: nC4 = … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/23399"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=23399"}],"version-history":[{"count":4,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/23399\/revisions"}],"predecessor-version":[{"id":23489,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/23399\/revisions\/23489"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=23399"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=23399"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=23399"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}