{"id":35469,"date":"2022-09-19T16:50:40","date_gmt":"2022-09-19T11:20:40","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35469"},"modified":"2022-09-19T16:50:40","modified_gmt":"2022-09-19T11:20:40","slug":"inter-2nd-year-maths-2a-permutations-and-combinations-solutions-ex-5b","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-2nd-year-maths-2a-permutations-and-combinations-solutions-ex-5b\/","title":{"rendered":"Inter 2nd Year Maths 2A Permutations and Combinations Solutions Ex 5(b)"},"content":{"rendered":"

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

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

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

Question 1.
\nFind the number of 4-digited numbers that can be formed using the digits 1, 2, 4, 5, 7, 8 when repetition is allowed.
\nSolution:
\nThe number of 4 digited numbers that can be formed using the digits 1, 2, 4, 5, 7, 8 when repetition is allowed = 64<\/sup> = 1296
\n\"Inter<\/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:
\nThe 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 = 55<\/sup> = 3125
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 3.
\nFind the number of functions from a set A containing 5 elements into a set B containing 4 elements.
\nSolution:
\nSet A contains 5 elements.
\nSet 8 contains 4 elements.
\nHint: The total number of functions from set A containing m elements to set B containing n elements is nm<\/sup>.
\nFor the image of each of the 5 elements of the set, A has 4 choices.
\n\u2234 The number of functions from set A containing 5 elements into a set B containing 4 elements = 4 \u00d7 4 \u00d7 4 \u00d7 4 \u00d7 4 (5 times)
\n= 45<\/sup>
\n= 1024<\/p>\n

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

Question 1.
\nFind the number of palindromes with 6 digits that can be formed using the digits
\n(i) 0, 2, 4, 6, 8
\n(iii) 1, 3, 5, 7, 9
\nSolution:
\nPalindromes mean first digit, sixth digit and second digit, fifth digit and third digit, and the fourth digit are the same numbers.
\nThat is we have filled the first three digits only and then the remaining digits are the same.
\n\"Inter
\n(i) Given numbers are 0, 2, 4, 6, 8
\nFor the required 6-digit palindromes first digit can be filled 4 ways except ‘0’.
\nThe second digit can be filled in 5 ways and the third can be filled in 5 ways.
\n\u2234 The number of 6-digit palindromes using the digits 0, 2, 4, 6, 8 are 4 \u00d7 52<\/sup> = 100
\n\"Inter
\n(ii) Given numbers are 1, 3, 5, 7, 9
\nFor the required 6-digit palindromes first digit can be filled in 5 ways, the second digit can be filled in 5 ways and the third digit can be filled in 5 ways.
\n\u2234 The number of 6-digit palindromes using the digits 1, 3, 5, 7, 9 are 53<\/sup> = 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:
\nThe number of 4 digited numbers formed using the digits 1, 2, 3, 4, 5, 6 when repetition is allowed = 64<\/sup>
\nThe number of 4 digited numbers formed using the digits 1, 2, 3, 4, 5, 6 when repetition is not allowed = 6<\/sup>P4<\/sub>
\nThe number of 4 digited telephone numbers in which atleast one digit is repeated = 64<\/sup> – 6<\/sup>P4<\/sub>
\n= 64<\/sup> – 6 \u00d7 5 \u00d7 4 \u00d7 3
\n= 1296 – 360
\n= 936<\/p>\n

\"Inter<\/p>\n

Question 3.
\nFind the number of bijections from a set A containing 7 elements onto itself.
\nSolution:
\nHint: The number of bijections from set A with n elements to set B with the same number of elements in A is n!
\nLet A = [a1<\/sub>, a2<\/sub>, a3<\/sub>, a4<\/sub>, a5<\/sub>, a6<\/sub>, a7<\/sub>]
\nFor a bijection, the element a1<\/sub> has 7 choices
\nfor its image, the element a2<\/sub> has 6 choices
\nfor its image, the element a3<\/sub> has 5 choices
\nfor its image, the element a4<\/sub> has 4 choices
\nfor its image, the element a5<\/sub> has 3 choices
\nfor its image, the element a6<\/sub> has 2 choices
\nfor its image and the element, a7<\/sub> has 1 choice for its image.
\n\u2234 The number of bijections from set A with 7 elements onto itself = 7!
\n= 7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1
\n= 5040<\/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
\n(i) when repetition is allowed is nr<\/sup>
\n(ii) when repetition is not allowed is n<\/sup>Pr<\/sub>
\n\u2234 The number of ways of arranging ‘r’ things in a line using the \u2018n1 different things in which atleast one thing is repeated = nr<\/sup> – n<\/sup>Pr<\/sub><\/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:
\nFirst, we can fill up the first place with N in one way.
\n\"Inter
\nThe remaining 4 places can be filled with any one of the 6 letters in 6 \u00d7 6 \u00d7 6 \u00d7 6 = 64<\/sup> ways.
\n\u2234 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 = 1 \u00d7 64<\/sup> = 1296<\/p>\n

\"Inter<\/p>\n

Question 6.
\nFind the number of 5-digit numbers divisible by 5 that can be formed using the digits 0, 1, 2, 3, 4, 5 when repetition is allowed.
\nSolution:
\nThe unit place of 5 digited numbers which can be divisible by 5 using the given digits can be filled by either 0 or 5 in two ways.
\nThe first place can be filled in any one of the given digits except ‘0’ in 5 ways.
\nThe remaining 3 places can be filled by any one of the given digits in 6 \u00d7 6 \u00d7 6 ways (\u2235 repetition is allowed)
\n\"Inter
\n\u2234 The number of 5 digited numbers divisible by 5 that can be formed using the given digits when repetition is allowed = 2 \u00d7 5 \u00d7 6 \u00d7 6 \u00d7 6 = 2160 ways.<\/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:
\nAll the single digited numbers, two digited numbers, three digited numbers and the four digited numbers started with 1 are the numbers less than 2000 using the digits 1, 2, 3, 4.
\nThe number of single digited numbers formed using the given digits = 4
\n\"Inter
\nThe number of two digited numbers formed using the given digits when repetition is allowed = 4 \u00d7 4 = 16
\n\"Inter
\nThe number of three digited numbers formed using the given digits = 4 \u00d7 4 \u00d7 4 = 64
\n\"Inter
\nThe number of 4 digited numbers started with 1 formed using the given digits = 4 \u00d7 4 \u00d7 4 = 64
\n\"Inter
\n\u2234 The total number of numbers less than 2000 that can be formed using the digits 1, 2, 3, 4 if repetition is allowed = 4 + 16 + 64 + 64 = 148<\/p>\n

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

Question 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 have
\n(i) no letter is repeated
\n(ii) At atleast one letter is repeated
\nSolution:
\nThe number of 4 letter words can be formed using the 9 different letters of an alphabet when repetition is allowed = 94<\/sup>
\n(i) The number of 4 letter words can be formed using the 9 different letters of an alphabet in which no letter is repeated = 9<\/sup>P4<\/sub>
\n= 9 \u00d7 8 \u00d7 7 \u00d7 6
\n= 3024
\n(ii) The number of 4 letter words can be formed using the 9 different letters of an alphabet in which atleast one letter is repeated = 94<\/sup> – 9<\/sup>P4<\/sub>
\n= 6561 – 3024
\n= 3537<\/p>\n

\"Inter<\/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 (i) 2 (ii) 4 when repetition is allowed.
\nSolution:
\n(i) First place can be filled by either 2 or 5 or 7 or 8 in 4 ways.
\n\"Inter
\nSecond place can be filled by any one of the given digits in 5 ways.
\nThird place can be filled by any one of the given digits in 5 ways.
\nLast place (or units place) can be filled by either 0 or 2 or 8 in 3 ways.
\n\u2234 The number of 4 digited divisible by 2 numbers that can be formed using the digits 0, 2, 5, 7, 8 when repetition is allowed = 4 \u00d7 5 \u00d7 5 \u00d7 3 = 300
\n(ii) Since a number is divisible by 4, the last two places should be filled with one of the 00, 08, 20, 28, 52, 72, 80, 88 in 8 ways.
\nThe first place can be filled 4 ways except 0. Second place can be filled in 5 ways.<\/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:
\nSince a number is divisible by 4, the last two places should be filled with one of the 00, 04, 12, 20, 24, 32, 40, 44 = 8 ways
\n\"Inter
\nThe first place can be filled in any one of the given digits except ‘0’ in 4 ways. The remaining 2 places can be filled by anyone the given digits in 5 \u00d7 5 ways.
\n\u2234 The number of 5 digited numbers that can be formed using the digits 0, 1, 2, 3, 4 that are divisible by 4 when repetition is allowed = 4 \u00d7 52<\/sup> \u00d7 8 = 800<\/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(b) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2A Permutations and Combinations Solutions Exercise 5(b) I. Question 1. Find the number of 4-digited numbers that can be formed using the digits …<\/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.guru\/wp-json\/wp\/v2\/posts\/35469"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35469"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35469\/revisions"}],"predecessor-version":[{"id":35470,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35469\/revisions\/35470"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35469"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35469"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35469"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}