{"id":9434,"date":"2023-11-02T16:36:54","date_gmt":"2023-11-02T11:06:54","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=9434"},"modified":"2023-11-03T16:59:52","modified_gmt":"2023-11-03T11:29:52","slug":"ts-inter-2nd-year-maths-2a-solutions-chapter-5-ex-5b","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2a-solutions-chapter-5-ex-5b\/","title":{"rendered":"TS Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(b)"},"content":{"rendered":"

Students must practice this TS Intermediate Maths 2A Solutions<\/a> Chapter 5 Permutations and Combinations Ex 5(b) 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(b)<\/h2>\n

I.
\nQuestion 1.
\nFind the number of 4 – digit numbers that can be formed using the digits 1, 2, 4, 5, 7, \u2022 8 when repetition is allowed.
\nSolution:
\nGiven digits are 1, 2, 4, 5, 7, 8.
\nAs repetitions are allowed,
\nEach place of 4 – digit number can be filled by given ‘6’ digits in 6 ways.
\n\u2234 By fundamental principle of counting number of 4 – digit numbers are 6 \u00d7 6 \u00d7 6 \u00d7 6 = 64<\/sup> = 1296.<\/p>\n

Question 2.
\nFind the number of 5 letter words that can be formed using the letters of the word RHYME if each letter can be used any number of times.
\nSolution:
\nGiven word RHYME contains 5 different letters.
\nAs repetitions are allowed, each blank of 5 letter words that can be formed using letters of word RHYME is 5 \u00d7 5 \u00d7 5 \u00d7 5 \u00d7 5 = 55<\/sup> = 3125.<\/p>\n

Question 3.
\nFind the number of functions from a set A containing 5 elements into a set B containing 4 elements.
\nSolution:
\nLet A = {a1<\/sub>, a2<\/sub>, a3<\/sub>, a4<\/sub>, a5<\/sub>} and B = {b1<\/sub>, b2<\/sub>, b3<\/sub>, b4<\/sub>}
\nTo define the image of a, we have 4 choices in set B.
\ni. e., Each element of set A has 4 choices in set B.
\n\u2234 The number of functions from A to B is 4 \u00d7 4 \u00d7 4 \u00d7 4 \u00d7 4 = 45<\/sup> = 1024.<\/p>\n

\"TS<\/p>\n

II.
\nQuestion 1.
\nFind the number of palindromes with 6 digits that can be formed using the digits
\ni) 0, 2, 4, 6, 8
\nii) 1, 3, 5, 7, 9.
\nSolution:
\ni) Given digits are 0, 2, 4, 6, 8.
\nThe first place and last place (lakh’s place and unit’s place) of a 6 – digit palindrome number is filled by same digit.
\nThis can be done in 4 ways (using non\u00aczero digit).
\nSimilarly Ten thousand\u2019s place and ten’s place is filled by same digit.
\nAs repetition is allowed this can be done in 5 ways.
\nThousand’s place and Hundred’s place is filled by same digit in 5 ways.
\n\u2234 Total number of 6 digital palindromes formed using given digits are 4 \u00d7 5 \u00d7 5 = 100.<\/p>\n

ii) Given digits are 1, 3, 5, 7, 9.
\nThe first place and last place (i.e., lakh’s place and unit’s place) of a 6 – digit palindrome number is filled by same digit in 5 ways.
\nAs repetitions allowed,
\nSimilarly ten thousand’s place & ten’s place is filled by same digit in 5 ways.
\nThousand’s place and Hundred’s place is filled by same digit in 5 ways.
\nTotal number of 6 digit palindrome formed using given digits are 5 \u00d7 5 \u00d7 5 = 125.<\/p>\n

Question 2.
\nFind the number of 4 – digit telephone numbers that can be formed using the digits 1, 2, 3, 4, 5, 6 with atleast one digit repeated.
\nSolution:
\nGiven digits are 1, 2, 3, 4, 5, 6.
\nThe number of 4-digit telephone numbers that can be formed using the given 6 digits.
\nCase – (i) :
\nWhen repetitions is allowed = 64<\/sup>
\nCase – (ii) :
\nWhen repetitions is not allowed
\nHence the number of 4 digit telephone numbers in which atleast one digit repeated is 64<\/sup> – \\({ }^6 \\mathrm{P}_4\\) = 936.<\/p>\n

Question 3.
\nFind the number of bijections from a set A containing 7 elements onto itself.
\nSolution:
\nLet A = {a7, a2 a7) i.e., set containing 7 elements.
\nThe bijection is both one-one and onto.
\nSo, to define the image of a1<\/sub> we have 7 choices.
\nThen we can define the image of a2<\/sub> is 6 ways.
\nSimilarly we can define the image of a3<\/sub> in 5 ways.
\nProceeding like this, the image of a7<\/sub> is defined only in one way.
\n\u2234 The number of bijections from A onto A is 7 \u00d7 6 \u00d7 5 \u00d7 …………… \u00d7 1 = 7! = 5040.<\/p>\n

\"TS<\/p>\n

Question 4.
\nFind the number of ways of arranging ‘r’ things in a line using the given ‘n’ different things in which atleast one thing is repeated.
\nSolution:
\nThe number of ways of arranging r’ things in a line using the given n’ different things, when repetitions is allowed is nr<\/sup>.
\nThe number of ways of arranging ‘r’ things in a line using the given n’ different things. When repetitions is not allowed is \\({ }^n P_r\\).
\n\u2234 The number of ways of arranging ‘r’ things in a line using n\u2019 different things so that atleast one thing is repeated is nr<\/sup> – \\({ }^n P_r\\).<\/p>\n

Question 5.
\nFind the number of 5 letter words that can be formed using the letters of the word NATURE that begin with N when repetition is allowed.
\nSolution:
\nGiven word is ‘NATURE’.
\nAs the 5 letter word begins with ‘N’, the first place is filled in only 1 way.
\nAs repetitions is allowed, each place of remaining 4 places can be filled in 6 ways.
\n\u2234 Total number of 5 letter words formed are 64<\/sup> = 1296.<\/p>\n

Question 6.
\nFind the number of 5-digit numbers divis-ible by 5 that can be formed using the digits 0, 1, 2, 3, 4, 5, when repetition is allowed.
\nSolution:
\nGiven digits are 0, 1, 2, 3, 4, 5.
\nThe ten thousand’s place of a 5 – digit numbers formed using given digits can be filled in 5 ways.
\nAs the 5 – digit number is divisible by ‘5’, the unit’s place can be filled in 2 ways.
\n(i.e., either 0 or 5).
\n\u2234 The remaining 3 places can be filled in 6 ways each.
\n\u2234 Number of 5 digit numbers divisible by 5 formed using given digits = 5 \u00d7 2 \u00d7 63<\/sup> = 2160.<\/p>\n

\"TS<\/p>\n

Question 7.
\nFind the number of numbers less than 2000 that can be formed using the digits, 1, 2, 3, 4 if repetition is allowed.
\nSolution:
\nGiven digits are 1, 2, 3, 4.
\n\u2234 Number of single digit numbers formed is 4.
\nNumber of 2 digit numbers formed when repetitions is allowed is 42<\/sup> = 16.
\nNumber of 3 – digit numbers formed when repetitions is allowed is 43<\/sup> = 64.
\nFor 4 – digit number less than 2000, the thousands place is filled in 1 way, remaining 3 places can be filled in 4 ways each.
\n\u2234 Number of 4 digit numbers = 43<\/sup> = 64.
\n\u2234 Total number of numbers less than 2000 using given digits = 4 + 16 + 64 + 64 = 148.<\/p>\n

III.
\nQuestion 1.
\n9 different letters of an alphabet are given. Find the number of 4 letter words that can be formed using these 9 letters which bave
\ni) no letter is repeated
\nii) atleast one letter is repeated.
\nSolution:
\nThe number of 4 letter words that can be formed using 9 different letters when repeti tion is allowed = 94<\/sup>
\ni) The number of 4 letter words that can be formed using 9 dIfferent letters when no letter is repeated = \\({ }^9 \\mathrm{P}_4\\)
\nii) The number of 4 letter words that can be formed using 9 different letters so that atleast one letter is repeated = 94<\/sup> – \\({ }^9 \\mathrm{P}_4\\) = 3537.<\/p>\n

Question 2.
\nFind the number of 4-digit numbers which can be formed using the digits 0, 2, 5, 7, 8 that are divisible by
\n(i) 2
\n(ii) 4 when repetition is allowed.
\nSolution:
\nGiven digits are 0, 2, 5, 7, 8.<\/p>\n

i) Divisible by 2:
\nThe thousand\u2019s place of 4 digit number when repetition is allowed can be filled in 4 ways. (using non-zero digits)
\nThe 4-digit number is divisible by 2, when the units place is an even digit. This can be done in 3 ways.
\nThe remaining 2 places can be filled by 5 ways each i.e., 52<\/sup> or 25 ways.
\n\u2234 Number of 4 digit numbers which are divisible by 2 is 4 \u00d7 3 \u00d7 25 = 300.<\/p>\n

ii) Divisible by 4:
\nA number is divisible by 4 only when the number in last two places (tens and units) is a multiple of 4.
\nAs repetition is allowed the last two places should be filled with one of the following 00, 08, 20, 28, 52, 72 80, 88
\nThis can be done is 8 ways.
\nThousands place is filled in 4 ways. (i.e., using non-zero digits)
\nHundreds place can be filled in 5 ways.
\n\u2234 Total number of 4 digit numbers formed = 8 \u00d7 4 \u00d7 5 = 160.<\/p>\n

Question 3.
\nFind the number of 4-digit numbers that can be formed using the digits 0, 1, 2, 3, 4, 5 which are divisible by 6 when repetition of the digits is allowed.
\nSolution:
\nGiven digits are 0, 1, 2, 3, 4, 5.
\nThousands place can be filled in 5 ways, (using non-zero digit) when repetition is allowed.
\nHundred\u2019s place and ten\u2019s place can be filled in 6 ways each, i.e., 62 ways.
\nIf we fill up the unit\u2019s place in 6 ways, we get 6 consecutive positive integers.
\nOut of any six consecutive integers only one is divisible by \u20186\u2019.
\nHence unit\u2019s place is filled in 1 way.
\nHence number of 4 digit numbers which are divisible by 6 using given digits = 5 \u00d7 62<\/sup> \u00d7 1 = 180.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice this TS Intermediate Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(b) to find a better approach to solving the problems. TS Inter 2nd Year Maths 2A Solutions Chapter 5 Permutations and Combinations Ex 5(b) I. Question 1. Find the number of 4 – digit numbers that can be formed using … 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\/9434"}],"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=9434"}],"version-history":[{"count":3,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9434\/revisions"}],"predecessor-version":[{"id":9437,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/9434\/revisions\/9437"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=9434"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=9434"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=9434"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}