{"id":9464,"date":"2023-11-02T17:00:48","date_gmt":"2023-11-02T11:30:48","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=9464"},"modified":"2023-11-03T17:00:13","modified_gmt":"2023-11-03T11:30:13","slug":"ts-inter-2nd-year-maths-2a-solutions-chapter-5-ex-5c","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2a-solutions-chapter-5-ex-5c\/","title":{"rendered":"TS Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(c)"},"content":{"rendered":"

Students must practice this TS Intermediate Maths 2A Solutions<\/a> Chapter 5 Permutations and Combinations Ex 5(c) to find a better approach to solving the problems.<\/p>\n

TS Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(c)<\/h2>\n

I.
\nQuestion 1.
\nFind the number of ways of arranging 7 persons around a circle.
\nSolution:
\nWe know that number of circular permutations of \u2018n\u2019 dissimilar things (taken all at atime) is (n – 1)!. The number of ways ofarranging 7 persons, around a circle is (7 – 1)! = 6! = 720.<\/p>\n

Question 2.
\nFind the number of ways of arranging the chief minister and 10 cabinet ministers at a circular table so that the chief minister always sits in a particular seat.
\nSolution:
\nThe chief minister always sits in a particular seat, hence, he is arranged in only 1 way.
\nNow the 10 cabinet ministers in 10 places are arranged in 10! ways.
\n\u2234 Total number of ways 1 \u00d7 10! = 10!<\/p>\n

Question 3.
\nFind the number of ways of preparing a chain with 6 different coloured beads.
\nSolution:
\nWe know that the number of circular permutations of hanging type that can be formed using n things is \\(\\frac{(n-1) !}{2}\\).
\nHence the number of different ways of preparing the chains with 6 different coloured beads = \\(\\frac{(6-1) !}{2}=\\frac{5 !}{2}\\) = 60.<\/p>\n

\"TS<\/p>\n

II.
\nQuestion 1.
\nFind the number of ways of arranging 4 boys and 3 girls around a circle so that all the girls sit together.
\nSolution:
\nTreat 3 girls as 1 unit.
\nThis unit along with 4 boys becomes 5 entities.
\nNumber of circular permutations of these units = (5 – 1) ! = 4! = 24
\nThree girls can be arranged themselves in 3! ways.
\n\u2234 Total number of ways = 24 \u00d7 3! = 144.<\/p>\n

Question 2.
\nFind the number of ways of arranging 7 gents and 4 ladies around a circular table if no two ladies wish to sit together.
\nSolution:
\nAs no two ladies wish to sit together, first we arrange 7 gents.
\nThese 7 gents around a circular table can be arranged in (7 – 1) ! ways i.e., 6! ways.
\nNow the number of gaps formed are 7.
\nNumber of ways of arranging 4 ladies in these 7 gaps = \\({ }^7 \\mathrm{P}_4\\).
\n\u2234 Total number of ways of arranging 7 gents and 4 ladies around a circular table if no two ladies wish to sit together = 6! \u00d7 \\({ }^7 \\mathrm{P}_4\\).<\/p>\n

Question 3.
\nFind the number of ways of arranging 7 guests and a host around a circle if 2 par-ticular guests wish to sit on either side of the host.
\nSolution:
\nNumber of guests are 7.
\nTreat 2 particular guests and host as single unit.
\nThis unit with remaining 5 guests becomes 6 entities.
\n\u2234 Number of ways of arranging 6 entities around a circle = (6 – 1) ! = 5!
\nThe 2 particular guest can arrange on either side of the host in 2! ways.
\n\u2234 Number of ways of arranging = 5! \u00d7 2! = 240.<\/p>\n

\"TS<\/p>\n

Question 4.
\nFind the number of ways of preparing a garland with 3 yellow, 4 white and 2 red roses of different sizes such that the two red roses come together.
\nSolution:
\nTreat 2 red roses of different sizes as single unit, which can be arranged in 2! ways. This unit with 3 yellow and 4 white roses of different sizes becomes 8 entities.
\nThe number of ways of preparing a garland with 8 entities are (8 – 1)! = 7! ways.
\n\u2234 Number of circular permutations are 7! \u00d7 2!
\nBut this being the case of garland, clockwise and anti clockwise arrangements look alike. Hence the required number of ways = \\(\\frac{1}{24}\\) \u00d7 7! \u00d7 2! = 5040.<\/p>\n

III.
\nQuestion 1.
\nFind the number of ways of arranging 6 boys and 6 girls around a circular fable so that
\ni) all the girls sit together
\nii) no two girls sit together
\niii) boys and girls sit alternately.
\nSolution:
\nGiven 6 boys and 6 girls.
\ni) All the girls sit together :
\nTreat all the girls as 1 unit. Then we have 6 boys and 1 unit of girls.
\nThey can be arranged around a circular table in 6! ways.
\nAgain, the 6 girls can be arranged themselves in 6! ways.
\nTotal number of arrangements = 6! \u00d7 6!.<\/p>\n

ii) No two girls sit together :
\nAs no two girls sit together, first we arrange 6 boys around a circular table.
\nThis can be done in 5! ways.
\nThen we can find 6 gaps between them.
\n6 girls in these 6 gaps can be arranged in 6! ways.
\n\u2234 The number of arrangements = 5! \u00d7 6!.<\/p>\n

iii) Boys and girls sit alternately :
\nAs number of boys is equal to number of girls, the arrangement of boys and girls sit alternately is same as no two girls sit together.
\nFirst arrange 6 boys around circular table.
\nThis can be done in 5! ways.
\nThen we find 6 gaps.
\nArranging 6 girls in these 6 gaps can be done in 6! ways.
\n\u2234 Total number of arrangements = 5! \u00d7 6!.<\/p>\n

\"TS<\/p>\n

Question 2.
\nFind the number of ways of arranging 6 red roses and 3 yellow roses of different sizes into a garland. In how many of them
\ni) all the yellow roses are together
\nii) no two yellow roses are together.
\nSolution:
\nGiven 6 red roses and 3 yellow roses of different sizes.
\n\u2234 Total number of roses are 9.
\n\u2234 The number of ways of arranging 6 red roses and 3 yellow roses of different sizes into a garland = \\(\\frac{(9-1) !}{2}=\\frac{8 !}{2}\\) = 20160.<\/p>\n

i) All the yellow roses are together :
\nTreat yellow roses as one unit.
\nThen this unit with 6 red roses can have the circular permutations in (7 – 1)! = 6! ways.
\nNow 3 yellow roses can be arranged themselves in 3! ways.
\nBut in the case of garlands, clockwise and anti clockwise arrangements look alike.
\n\u2234 The number of arrangements = \\(\\frac{6 ! \\times 3 !}{2}\\) = 2160.<\/p>\n

ii) No two yellow roses are together :
\nAs no two yellow roses are together, first arrange 6 red roses in garland form.
\nThis can be done in 5! ways.
\nThen we find 6 gaps. Arrangement of 3 yellow roses in these 6 gaps can be done in \\({ }^6 P_3\\) ways.
\nBut in the case of garlands, clockwise and anti clockwise arrangements look alike.
\n\u2234 The number of arrangements = \\(\\frac{1}{2} \\times 5 ! \\times{ }^6 P_3\\) = 7200.
\n3 Chinese can be arranged themselves in 3! ways.
\n3 Canadians can be arranged themselves in 3! ways.
\n2 Americans can be arranged themselves in 2! ways.
\n\u2234 The number of required arrangements are 3! \u00d7 3! \u00d7 3! \u00d7 3! \u00d7 2! = 2592.<\/p>\n

\"TS<\/p>\n

Question 4.
\nA chain of beads is to be prepared using 6 different red coloured beads and 3 differ-ent blue coloured beads. In how many ways can this be done so that no two blue coloured beads come together.
\nSolution:
\nGiven 6 different red coloured beads and 3 different blue coloured beads.
\nAs no two blue coloured beads come together, first arrange 6 red coloured beads in the form of chain.
\nThis can be done in (6 – 1)! = 5! ways.
\nThen 6 gaps are formed between them.
\nNow arrangement of 3 different blue coloured beads in these 6 gaps can be done in hP3 ways.
\nThen total number of circular permutations are \\({ }^6 \\mathrm{P}_3\\) \u00d7 5!.
\nBut, this being the case of chain, clockwise and anti clockwise look alike.
\nHence required number of ways = \\(\\frac{1}{2} \\times{ }^6 \\mathrm{P}_3 \\times 5\\) = 7200.<\/p>\n

Question 5.
\nA family consists of father, mother, 2 daugh\u00acters and 2 sons. In how many different ways can they sit at a round table if the 2 daughters wish to sit on either side of the father ?
\nSolution:
\nTreat 2 daughters and father as 1 unit.
\nThis unit with mother and 2 sons becomes 4 entities.
\nNumber of ways can 4 entities arranged around circular table are (4 – 1) ! = 3! ways.
\nTwo daughters on either side of the father can be arranged in 2! ways.
\n\u2234 Required number of arrangements = 2! \u00d7 3! = 12.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice this TS Intermediate Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(c) to find a better approach to solving the problems. TS Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(c) I. Question 1. Find the number of ways of arranging 7 persons around a circle. Solution: … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9464"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=9464"}],"version-history":[{"count":2,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9464\/revisions"}],"predecessor-version":[{"id":9650,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9464\/revisions\/9650"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=9464"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=9464"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=9464"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}