{"id":36726,"date":"2022-11-30T15:51:53","date_gmt":"2022-11-30T10:21:53","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=36726"},"modified":"2022-12-01T14:30:01","modified_gmt":"2022-12-01T09:00:01","slug":"maths-2a-permutations-and-combinations-important-questions-short-answer-type","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/maths-2a-permutations-and-combinations-important-questions-short-answer-type\/","title":{"rendered":"TS Inter Second Year Maths 2A Permutations and Combinations Important Questions Short Answer Type"},"content":{"rendered":"

Students must practice these Maths 2A Important Questions<\/a> TS Inter Second Year Maths 2A Permutations and Combinations Important Questions Short Answer Type to help strengthen their preparations for exams.<\/p>\n

TS Inter Second Year Maths 2A Permutations and Combinations Important Questions Short Answer Type<\/h2>\n

Question 1.
\nIf the letters of the word PRlSON are permuted in all possible ways and the words thus formed are arranged in dictionary order, find the rank of the word
\ni) PRISON.
\nii) SIPRON [Mar. \u201814, May\u2019 13, AP – Mar. 2017]
\nSolution:
\ni) The alphabetical order of the letters of the given word is I, N, O, P, R, S.<\/p>\n

\"TS<\/p>\n

The no. of words begin with I is 5! = 120
\nThe no. of words begin with N is 5! = 120
\nThe no. of words begin with O is 5! = 120
\nThe no. of words begin with Pl is 4! = 24
\nThe no. of words begin with PN is 4! = 24
\nThe no. of words begin with PN is 4! = 24
\nThe no. of words begin with PRlN is 2! = 2
\nThe no, of words begin with PRIO is 2! = 2
\nThe no. of words begin with PRISN is 1! = 1
\nThe next word is PRISON = 1
\n\u2234 Rank of the word, PRISON = 120 + 120 + 120 + 24 + 24 + 24 + 2 + 2 + 1 + 1 = 438.<\/p>\n

ii) The alphabetical order of the letters of the given word is I, N, O, P, R, S.<\/p>\n

\"TS<\/p>\n

The no. of words begin with L is 5! = 120
\nThe no. of words begin with N is 5! = 120
\nThe no. of words begin with Ois 5! = 120
\nThe no. of words begin with P is 5! = 120
\nThe no. of words begin with R is 5! = 120
\nThe no. of words begin with SIN is 3! 6
\nThe no. ol words begin with SlO is 3! = 6
\nThe no. of words begin with SIPN is 2! = 2
\nThe no. of words begin with SIPO is 2! = 2
\nThe no. of wor(ls begin with SIPRN is 1! = 1
\nThe next word is SIPRON = 1
\n\u2234 The Rank of SIPRON is 120 + 120 + 120 + 120 + 120 + 6 + 6 + 2 + 2 + 1 + 1 = 618.<\/p>\n

\"TS<\/p>\n

Question 2.
\nIf the letters of the word MASTER are permuted In all possible ways and the words thus formed are arranged in the dictionary order, then find the ranks of the word REMAST. [March \u201808, Board Paper].
\nSolution:<\/p>\n

\"TS<\/p>\n

The alphabetical order of the letters of the given word is A, E. M, R, S, T.
\nThe no. of words begin with A is 5! = 120
\nThe no. of words begin with E is 5! = 120
\nThe no. of words begin with M is 5! = 120
\nThe no. of words begin with RA is 4! = 24
\nThe no. of words begin with REA is 3! = 6
\nThe next word is REMAST = 1
\n\u2234 The Rank of the word, REMAST = 120 + 120 + 120 + 246 + 1 = 391.<\/p>\n

Question 3.
\nIf the letters of the word MASTER are permuted in all possible ways and the words thus fonned are arranged in the dictionary order, then find the ranks f the word MASTER. [AP & TS – Mar. 2019] [May \u201811, \u201810, \u201807,\u201906; March \u201811,\u201909, \u201808, TS – Mar. 2016; AP – Mar., May 2015].
\nSolution:<\/p>\n

\"TS<\/p>\n

The alphabetical order of the letters of thi given word is A, E, M, R, S, T.
\nThe no. of words begin with A is 5! = 120
\nThe no. of words begin with E is 5! = 120
\nThe no. of words begin with MAE is 3! = 6
\nThe no. of words begin with MAR is 3! = 6
\nThe no. of words begin with MASE is 2! = 2
\nThe no. of words begin with MASR is 2! = 2
\nThe next word is MASTER = 1
\nThe Rank of the word, MASTER = 120 + 120 + 6 + 6 + 2 + 2 + 1 = 257.<\/p>\n

Question 4.
\nFind the sum of all 4-digit numbers that can be formed using the digits 1, 3, 5, 7, 9. [AP & TS – Mar. 18; May \u201814, Mar.13]
\nSolution:
\nThe no. of four digited number formed by using the digits 1, 3, 5, 7, 9 without repetition = \\({ }^n P_r={ }^5 P_4\\)
\n= 5 . 4. 3 . 2 = 120.
\nOut of these 120 numbers, \\({ }^n P_r={ }^5 P_4\\) number contain 1 in units place.
\n\\({ }^4 \\mathrm{P}_3\\) numbers contain 1 in Ten’s place.
\n\\({ }^4 \\mathrm{P}_3\\) numbers contain 1 in hundred’s place.
\n\\({ }^4 \\mathrm{P}_3\\) numbers contain 1 in thousand’s place.<\/p>\n

\"TS<\/p>\n

\u2234 The value obtained by adding (1) in all the numbers = \\({ }^4 \\mathrm{P}_3\\) \u00d7 1 \u00d7 1 + \\({ }^4 \\mathrm{P}_3\\) \u00d7 1 \u00d7 10 + \\({ }^4 \\mathrm{P}_3\\) \u00d7 1 \u00d7 100 + \\({ }^4 \\mathrm{P}_3\\) \u00d7 1 \u00d7 1000
\n= \\({ }^4 \\mathrm{P}_3\\) \u00d7 1 + \\({ }^4 \\mathrm{P}_3\\) \u00d7 10 + \\({ }^4 \\mathrm{P}_3\\) \u00d7 100 + \\({ }^4 \\mathrm{P}_3\\) \u00d7 1000
\n= \\({ }^4 \\mathrm{P}_3\\) \u00d7 1 [1 + 10 + 100 + 1000]
\n= \\({ }^4 \\mathrm{P}_3\\) \u00d7 3 \u00d7 (1111)<\/p>\n

Similarly the value obtained by adding (3) is \\({ }^4 \\mathrm{P}_3\\) \u00d7 3 \u00d7 1111
\nThe value obtained by adding (5) is \\({ }^4 \\mathrm{P}_3\\) \u00d7 5 \u00d7 1111
\nThe value obtained by adding (7) is \\({ }^4 \\mathrm{P}_3\\) \u00d7 7 \u00d7 1111
\nThe value obtained by adding (9) is \\({ }^4 \\mathrm{P}_3\\) \u00d7 9 \u00d7 1111
\nThe sum of all the numbers = \\({ }^4 \\mathrm{P}_3\\) \u00d7 1 \u00d7 (1111) + \\({ }^4 \\mathrm{P}_3\\) \u00d7 3 \u00d7 (1111) + \\({ }^4 \\mathrm{P}_3\\) \u00d7 5 \u00d7 (1111) + \\({ }^4 \\mathrm{P}_3\\) \u00d7 7 \u00d7 (1111) + \\({ }^4 \\mathrm{P}_3\\) \u00d7 9 \u00d7 (1111)
\n= \\({ }^4 \\mathrm{P}_3\\) (1111) [1 + 3 + 5 + 7 + 9]
\n= 24 (1111) (25) = 6,66,600.<\/p>\n

\"TS<\/p>\n

Question 5.
\nFind the sum of aIl 4 digited numbers that can be formed using the digits 1, 2, 4, 5, 6 without repetition. [March ’10]
\nSolution:
\nThe number of 4 digited numbers formed by using the digits 1, 2, 4, 5, 6 without repetition = \\({ }^5 \\mathrm{P}_4\\)
\n= 5 . 4 . 3 . 2 = 120
\nOut of these 120 numbers,
\n\\({ }^4 \\mathrm{P}_3\\) numbers contain 1 in units places
\n\\({ }^4 \\mathrm{P}_3\\) numbers contain 1 in ten’s place
\n\\({ }^4 \\mathrm{P}_3\\) numbers contain 1 in hundred’s place
\n\\({ }^4 \\mathrm{P}_3\\) numbers contain 1 in thousand’s place<\/p>\n

\"TS<\/p>\n

\u2234 The value obtained by adding (1) in all the numbers
\n= \\({ }^4 \\mathrm{P}_3\\) \u00d7 1 \u00d7 1 + \\({ }^4 \\mathrm{P}_3\\) \u00d7 1 \u00d7 10 + \\({ }^4 \\mathrm{P}_3\\) \u00d7 1 \u00d7 100 + \\({ }^4 \\mathrm{P}_3\\) \u00d7 1 \u00d7 1000
\n= \\({ }^4 \\mathrm{P}_3\\) \u00d7 1 [1 + 10 + 100 + 1000]
\n= \\({ }^4 \\mathrm{P}_3\\) \u00d7 1 \u00d7 1111
\nSimilarly the value obtained by adding (2) is \\({ }^4 \\mathrm{P}_3\\) \u00d7 2 (1111)
\nThe value obtained by adding (4) is \\({ }^4 \\mathrm{P}_3\\) \u00d7 4 (1111)
\nThe value obtained by adding (5) is \\({ }^4 \\mathrm{P}_3\\) \u00d7 5 (1111)
\nThe value obtained by adding (6) is \\({ }^4 \\mathrm{P}_3\\) \u00d7 6 (1111)
\n\u2234 The sum of all the numbers = \\({ }^4 \\mathrm{P}_3\\) x 1×1111 + \\({ }^4 \\mathrm{P}_3\\) x2(1111) + \\({ }^4 \\mathrm{P}_3\\) x4(1111) + \\({ }^4 \\mathrm{P}_3\\) x5(1111)+ \\({ }^4 \\mathrm{P}_3\\)x6(1111)
\n= \\({ }^4 \\mathrm{P}_3\\) (1111) + (1 + 2 + 4 + 5 + 6)
\n= \\({ }^4 \\mathrm{P}_3\\) (1111) (18)
\n= 4 . 3 . 2 (1111) (18)
\n= 24 (1111) 18
\n= 479952.<\/p>\n

Question 6.
\nFind the number of numbers that are greater than 4000 which can be formed using the digits 0, 2, 4,6, 8 without repetition. [May \u201809]
\nSolution:
\nFrom the given digits, 0 cannot be placed in the first place. So, 1st place can be filled with the remaining four digits.
\nAll the numbers of five digits are greater than 4000.
\n4 digit numbers:<\/p>\n

\"TS<\/p>\n

In four digit numbers the number starting 4 with 4 or 6 or 8 are greater than 4000.
\nThe number of 4 digit numbers which begin with 4 or 6 or 8 = 3 . \\({ }^4 \\mathrm{P}_3\\)
\n= 3 . 4 . 3 . 2 = 72<\/p>\n

5 digit numbers:<\/p>\n

\"TS<\/p>\n

The number of five digit numbers = 4 . \\({ }^4 \\mathrm{P}_4\\)
\n= 4 . 4 . 3 . 2 . 1 = 96
\n\u2234 The number of numbers greater than 4000 = 72 + 96 = 168.<\/p>\n

Question 7.
\nFind the number of ways of arranging 5 different mathematics hooks, 4 different physics books and 3 different chemistry books such that the books of the same subject are together.
\nSolution:<\/p>\n

\"TS<\/p>\n

The number of ways of arranging three bundles of Maths, Physics and Chemistry books is \\({ }^3 \\mathrm{P}_3\\) = 3!.
\n5 different Mathematics books in the bundle are arranged among themselves in 5! ways.
\n4 different Physics books in the bundle are arranged among themselves in 4! ways.
\n3 different Chemistry books in the bundle are arranged among themselves in 3! ways.
\n\u2234 The required number of arrangements = 3! . 5! . 4! . 3!
\n= 6 . 120 . 24 . 6 = 103684.<\/p>\n

\"TS<\/p>\n

Question 8.
\nFind the number of ways of arranging 7 gents and 4 ladies around a circular table if no two ladles wish to sit together. [May \u201807]
\nSolution:
\nFirst arrange 7 gents around a circle in (7 – 1)! = 6! ways.<\/p>\n

\"TS<\/p>\n

Now arrange 4 ladies in 7 positions in \\({ }^7 \\mathrm{P}_4\\) ways.
\n\u2234 The required number of arrangements = 6! . \\({ }^7 \\mathrm{P}_4\\)
\n= 6 . 5 . 4 . 3 . 2 . 1 . 7 . 6 . 5 . 4 = 604800.<\/p>\n

Question 9.
\nFind the number of ways of arranging the letters of word SINGING so that
\ni) they begin and end with I
\nii) the two G\u2019s come together.
\nSolution:
\nThe given word has 7 letters \u00a1n which there are 21\u2019s are alike, 2N\u2019s are alike. 2G\u2019s are alike, 1S is different.
\ni) They begin and end with I:<\/p>\n

\"TS<\/p>\n

First we fill the first and last places with I\u2019s in \\(\\frac{2 !}{2 !}\\) ways.
\nNow, we fill the remaining 5 places with the remaining 5 letters in \\(\\frac{5 !}{2 ! 2 !}\\) ways.
\n\u2234 The number of required arrangments is \\(\\frac{5 !}{2 ! 2 !} \\cdot \\frac{2 !}{2 !}=\\frac{5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1}{2 \\cdot 2}\\) = 30.<\/p>\n

ii) The 2 G\u2019s come together:
\nTreat the 2Gs as one unit.
\nThen we have 6 letters in which there are 2I’s and 2N\u2019s.
\nThey can be arranged in \\(\\frac{6 !}{2 ! 2 !}\\) ways.
\nNow, the 2G’s can be arranged among themselves in \\(\\frac{2 !}{2 !}\\).
\n\u2234 The number of required arrangments is \\(\\frac{6 !}{2 ! 2 !} \\cdot \\frac{2 !}{2 !}\\) = 180.<\/p>\n

Question 10.
\nIf the letters of the word EAMCET are permuted in all possible ways and 1f the words they formed are arranged In the dictionary order, find the rank of the word EAMCET. [TS – May 2015; Mar. \u201812] [AP – May, Mar. 2016; TS – Mar. 2017, May ’16]
\nSolution:
\nGiven word EAMCET,
\nThe alphabetical order of the letters of the, ACEEMT.
\nThe no. of words begin with A = \\(\\frac{5 !}{2 !}\\) = 60
\nThe no. of words begin with C = \\(\\frac{5 !}{2 !}\\) = 60<\/p>\n

\"TS<\/p>\n

The no. of words begin with EAC = 3! = 6
\nThe no. of words begin with EAE = 3! = 6
\nThe next word is EAMCET = 1
\nRank of word EAMCET = 60 + 60 + 6 + 6 + 1 = 133.<\/p>\n

\"TS<\/p>\n

Question 11.
\nIf 1 \u2264 r \u2264 n,then show that \\({ }^n C_{r-1}+{ }^n C_r={ }^{(n+1)} C_r\\). [March\u201911, May \u20189]
\nSolution:<\/p>\n

\"TS<\/p>\n

Question 12.
\nProve that 3 \u2264 r \u2264 n, \\({ }^{(n-3)} C_r+3 \\cdot{ }^{(n-3)} C_{(r-1)}+3 \\cdot{ }^{(n-3)} C_{(r-2)}\\) + \\({ }^{(n-3)} C_{(r-3)}={ }^n C_r\\). [May ’12, ’09]
\nSolution:<\/p>\n

\"TS<\/p>\n

Question 13.
\nSimplify \\({ }^{34} C_5+\\sum_{r=0}^4(38-r) C_4\\). [AP – Mar. ’19, ’17, ’16; March ’12, May ’11].
\nSolution:<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 14.
\nFind the number of ways of selecting 3 girls and 3 boys out of 7 girls and 6 boys. [May ’09]
\nSolution:
\nThe no. of ways of selecting 3 girls and 3 boys out of 7 girls and 6 boys = \\({ }^7 \\mathrm{C}_3 \\cdot{ }^6 \\mathrm{C}_3\\)
\n= \\(\\frac{7 \\cdot 6 \\cdot 5}{1 \\cdot 2 \\cdot 3} \\cdot \\frac{6 \\cdot 5 \\cdot 4}{3 \\cdot 2 \\cdot 1}\\) = 700<\/p>\n

Question 15.
\nFind the number of ways of selecting 3 vowels and 2 consonants from the letters of the word EQUATION. [May ’11, March ’07]
\nSolution:
\nThe word EQUATION contains 5 vowels and 3 consonants.
\nThe 3 vowels can be selected from 5 vowels in \\({ }^5 \\mathrm{C}_3\\) ways.
\nThe 2 consonants can be selected from 3 consonants in \\({ }^3 \\mathrm{C}_2\\) ways.
\n\u2234 The required no. of ways of selecting 3 vowels and 2 consonants = \\({ }^5 \\mathrm{C}_3 \\cdot{ }^3 \\mathrm{C}_2\\)
\n= \\(\\frac{5 \\cdot 4 \\cdot 3}{3 \\cdot 2 \\cdot 1} \\cdot \\frac{3 \\cdot 2}{2 \\cdot 1}\\) = 30.<\/p>\n

Question 16.
\nProve that \\(\\frac{{ }^{4 n} C_{2 n}}{{ }^{2 n} C_n}=\\frac{1 \\cdot 3 \\cdot 5 \\ldots \\ldots(4 n-1)}{\\{1 \\cdot 3 \\cdot 5 \\ldots \\ldots(2 n-1)\\}^2}\\). [TS – Mar. ’19, May ’08, 2000, Mart. ’98] [AP – Mar. 2017, 2015, TS – May 2015].
\nSolution:<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 17.
\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. [May ’14, ’10, March ’05, Board Paper AP – May 2016; TS – Mar. 18, May 2016]
\nSolution:
\nTotal number of players in the team = 11.<\/p>\n

\"TS<\/p>\n

1) 6 Batsmen, 5 bowlers:
\nThe no. of ways of selecting a cricket team = \\({ }^7 \\mathrm{C}_6 \\cdot{ }^6 \\mathrm{C}_5\\)
\n= 7 . 6 = 42<\/p>\n

2) 5 Batsmen, 6 bowlers:
\nThe no. of ways of selecting cricket team = \\({ }^7 \\mathrm{C}_5 \\cdot{ }^6 \\mathrm{C}_6\\)
\n= \\(\\frac{7 \\cdot 6}{2 \\cdot 1}\\) . 1
\n= 7 . 3 = 21.<\/p>\n

Question 18.
\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. [TS – Mar. 2015; March \u201813. ’09, ’08]
\nSolution:
\nSince, the committee must contain always the Indians will be in majority, the members of the committee may be the following.<\/p>\n

\"TS<\/p>\n

i) 5 Indians, 0 Americans.
\nThe no. of ways of selecting committee = \\({ }^6 \\mathrm{C}_5 \\cdot{ }^5 \\mathrm{C}_0\\)
\n=6×1 = 6<\/p>\n

ii) 4 Indians, 1 American:
\nThe no. of ways of selecting committee = \\({ }^6 \\mathrm{C}_4 \\cdot{ }^5 \\mathrm{C}_1\\)
\n= \\(\\frac{6 \\cdot 5}{2 \\cdot 1}\\) . 5 = 75<\/p>\n

iii) 3 Indians, 2 Americans.
\nThe no. of ways of selecting committee = \\({ }^6 \\mathrm{C}_3 \\cdot{ }^5 \\mathrm{C}_2\\)
\n= \\(\\frac{6 \\cdot 5 \\cdot 4}{3 \\cdot 2 \\cdot 1} \\cdot \\frac{5 \\cdot 4}{2 \\cdot 1}\\) = 200
\nThe required no. of ways of selecting j committee = 6 + 75 + 200 = 281.<\/p>\n

\"TS<\/p>\n

Some More Maths 2A Permutations and Combinations Important Questions<\/h3>\n

Question 19.
\nIf \\({ }^n \\mathrm{P}_3\\) = 1320, find n. [May ’08, March ’10, ’07]
\nSolution:
\nGiven \\({ }^n \\mathrm{P}_3\\) = 1320 = 12 . 11 . 10
\n\\({ }^n P_3={ }^{12} P_3\\)
\n\u21d2 n = 12.<\/p>\n

Question 20.
\nIf \\((n+1) P_5:{ }^n P_6\\) = 2 : 7, find n.
\nSolution:
\nGiven \\(\\frac{(n+1)}{{ }^n P_5}=\\frac{2}{7}\\)
\n\\(\\frac{(n+1) n(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)(n-4)(n-5)}=\\frac{2}{7}\\)
\n\\(\\frac{(n+1)}{(n-4)(n-5)}=\\frac{2}{7}\\)
\n7n + 7 = 2 (n2<\/sup> – 5n – 4n + 20)
\n7n + 7 = 2n2<\/sup> – 18n + 40
\n2n2<\/sup> – 18n – 7n + 40 – 7 = 0
\n2n2<\/sup> – 25n + 33 = 0
\n2n2<\/sup> – 22n – 3n + 33 = 0
\n2n (n – 11) – 3 (n – 11) = 0
\n(n – 11) (2n – 3) = 0
\nn = 11; n = \\(\\frac{3}{2}\\)
\nSince, n is a integer then n = 11.<\/p>\n

Question 20.
\nIf \\({ }^{18} P_{(r-1)}:{ }^{17} P_{(r-1)}\\) = 9 : 7, find ‘r’.
\nSolution:
\nGiven, \\({ }^{18} \\mathrm{P}_{\\mathrm{r}-1}:{ }^{17} \\mathrm{P}_{\\mathrm{r}-1}\\) = 9 : 7<\/p>\n

\"TS<\/p>\n

Question 21.
\nFind the number of different chains that can be prepared using 7 dIfferent coloured beads.
\nSolution:
\nTotal number of beads n = 7
\nThe no. of different chains that can be prepared using 7 different coloured beads is = \\(\\frac{(7-1) !}{2}=\\frac{6 !}{2}\\) = 360.<\/p>\n

\"TS<\/p>\n

Question 22.
\nProve that \\({ }^{25} C_4+\\sum_{r=0}^4(29-r) C_3={ }^{30} C_4\\). [AP – Mar. 2018]
\nSolution:<\/p>\n

\"TS<\/p>\n

Question 23.
\nFind the number of ways of arranging 5 boys and 4 girls in a row so that the row begins and ends with boys.
\nSolution
\nThe total no. of persons is 9. (5 boys + 4 girls)<\/p>\n

\"TS<\/p>\n

Let us take 9 blank\u2019s,
\nFirst, we fill the first and last places with boys.
\nThis can be done in \\({ }^5 \\mathrm{P}_2\\) ways.
\nNow, we have to fill up the remaining 7 places with the remaining 7 persons (3 boys + 4 girls) In 7! ways.
\nThe required no. of arrangements is = \\({ }^5 \\mathrm{P}_2\\) . 7!
\n= 20 \u00d7 5040 = 100800.<\/p>\n

Question 24.
\nIn how many ways 9 mathematics papers can be arranged so that the best and the worst (i) may come together (ii) may not come together?
\nSolution:
\ni) If the best and worst papers are treated as one unit, then we have 9 – 2 + 1 = 7 + 1 = 8 papers.
\nNow these can be arranged in (7 + 1)! ways and the best and worst papers between themselves can be permuted in 2! ways.
\nTherefore the number of arrangements in which best and worst papers come twet her is 8! 2!.<\/p>\n

ii) Total number of ways of arranging 9 mathematics papers is 9!.
\nThe best and worst papers come together in 8! . 2! ways.
\nTherefore the number of ways they may not come together is
\n9! – 8! 2! = 8! (9 – 2) 8! \u00d7 7.<\/p>\n

\"TS<\/p>\n

Question 25.
\nFind the number of ways of arrangIng 6 boys and 6 girls \u00a1n a row. In how many of these arrangements
\ni) all the girls are together
\nii) no two girls are together
\niii) boys and girls come alternately?
\nSolution:
\n6 boys + 6 girls = 12 persons.
\nThey can be arranged in a row in (12)! ways.<\/p>\n

i) Treat the 6 girls as one unit. Then we have 6 boys + 1 unit of girls.
\nThey can be arranged in 7! ways.
\nNow, the 6 girls among themselves can be permuted in 6! ways.
\nHence, by the fundamental principle, the number of arrangements in which all 6 girls are together 7! x 6!.<\/p>\n

ii) First we arrange 6 boys in a row in 6! ways.
\nThe girls can be arranged in the 7 gaps between the boys.
\nThese gaps are shown below by the letter X.<\/p>\n

\"TS<\/p>\n

Now, the girls can be arranged in these 7 gaps in \\({ }^7 \\mathrm{P}_6\\) ways.
\nHence, by the fundamental principle, the number of arrange ments in which no two girls corner together is 6! \u00d7 7P6
\n= 6! \u00d7 7!
\n= 7 \u00d7 6! \u00d7 6!.<\/p>\n

iii) Let us take 12 places. The row may begin with either a boy or a girl. That is, 2 ways.
\nIf it begins with a boy, then all odd places (1, 3, 5, 7, 9, 11) will be occupied by boys and the even places (2, 4, 6, 8, 10, 12) by girls.
\nThe 6 boys can be arranged in the 6 odd places in 6! ways and the 6 girls can be arranged in the 6 even places in 6! ways.
\nThus the number of arrangements in which boys and girls come alternately is 2 x 6! x 6!.<\/p>\n

Note :
\nIn the above, one may think that questions (ii) and (iii) are same. But they are not (as evident from the answers).
\nIn Question (ii), after arranging 6 boys, we found 7 gaps and 6 girls are arranged in these 7 gaps.
\nHence one place remains vacant.
\nIt can be any one of the 7 gaps. But in Question (iii), the vacant place should either be at the beginning or at the ending but not in between.
\nThus, only 2 choices for the vacant place.<\/p>\n

Question 26.
\nFind the number of 4 letter words that can be formed using the letters of the word MIRACLE. How many of them
\ni) begin with an vowel
\nii) begin and end with vowels
\niii) end with a consonant ?
\nSolution:
\nThe word MIRACLE has 7 letters. Hence the number of 4 letter words that can be formed using these letters is \\({ }^7 \\mathrm{P}_4\\) = 7 \u00d7 6 \u00d7 5 \u00d7 4 = 840.
\nLet us take 4 blanks.<\/p>\n

\"TS<\/p>\n

i) We can fill the first place with one of the 3 vowels (I, A, E) in \\({ }^3 \\mathrm{P}_1\\) = 3 ways.
\nNow, the remaining 3 places can be filled using the re\u00acmaining 6 letters in \\({ }^6 \\mathrm{P}_3\\) = 120 ways.
\nThus the number of 4 letter words that begin with an vowel is 3 \u00d7 120 = 360.<\/p>\n

ii) Fill the first and last places with 2 vowels in \\({ }^3 \\mathrm{P}_2\\) = 6 ways.
\nThe remaining 2 places can be filled with the remaining 5 letters in \\({ }^5 \\mathrm{P}_2\\) = 20 ways.
\nThus the number of 4 letter words that begin and end with vowels is 6 \u00d7 20 = 120.<\/p>\n

iii) We can fill the last place with one of the 4 consonants (M, R, C, L) in \\({ }^4 \\mathrm{P}_1\\) = 4 ways.
\nThe remaining 3 places can be filled with the letters in \\({ }^6 \\mathrm{P}_3\\) ways.
\nThus the number of 4 letter words that end with an vowel is 4 \u00d7 \\({ }^6 \\mathrm{P}_3\\) = 4 \u00d7 120 = 480.<\/p>\n

\"TS<\/p>\n

Question 27.
\nFind the number of ways of permuting the letters of the word PICTURE so that
\ni) all vowels come together
\nii) no two vowels come together.
\niii) the relative positions of vowels and consonants are not disturbed.
\nSolution:
\nThe word PICTURE has 3 vowels (I, U, E) and 4 consonants (P, C, T, R).
\ni) Treat the 3 vowels as one unit. Then we can arrange 4 consonants + 1 unit of vowels in 5! ways. Now the 3 vowels among themselves can be permuted in 3! ways. Hence the number of per\u00acmutations in which the 3 vowels come together is 5! \u00d7 3! = 720.<\/p>\n

ii) First arrange the 4 consonants in 4! ways. Then in between the vowels, in the beginning and in the ending, there are 5 gaps as shown below by the letter X.<\/p>\n

\"TS<\/p>\n

In these 5 places we can arrange the 3 vowels \\({ }^5 \\mathrm{P}_3\\) ways.
\nThus the number of words in which no two vowels come together is 4! \u00d7 \\({ }^5 \\mathrm{P}_3\\)
\n= 24 \u00d7 60 = 1440.<\/p>\n

iii) The three vowels can be arranged in their relative positions in 3! ways and the 4 consonants can be arranged in their relative positions in 4! ways.<\/p>\n

\"TS<\/p>\n

The required number of arrangements is 3! . 4! = 144.<\/p>\n

Note :
\nIn the above problem, from (i) we get that the number of permutations in which the vowels do not come together is
\n= Total number of permutations – number of permutations in which 3 vowels come together.
\n= 7! – 5! . 3!
\n= 5040 – 720 = 4320.
\nBut the number of permutations in which no two vowels come together is only 1440.
\nIn the remaining 2880 permutations two vowels come together and third appears away from these.<\/p>\n

Question 28.
\nFind the number of 4-digit numbers that can be formed using the digits 2, 3, 5, 6, 8 (without repetition). How many of them are divisible by
\ni) 2
\nii) 3
\niii) 4
\niv) 5
\nv) 25
\nSolution:<\/p>\n

\"TS<\/p>\n

The number of 4 digit numbers that can be formed using the 5 digits 2, 3, 5, 6, 8 is \\({ }^5 \\mathrm{P}_4\\) = 120.<\/p>\n

i) divisible by 2 :<\/p>\n

\"TS<\/p>\n

A number is divisible by 2, when its unit place must be filled with an even digits (2 or 6 or 8) from among the given integers. This can be done in three ways.
\nNow, the remaining three places can be filled with remaining four digits in \\({ }^4 P_3\\) = 4 . 3 . 2 = 24 ways.
\n\u2234 The no. of 4 digited numbers divisible by 2 = 3 . 24 = 72<\/p>\n

\"TS<\/p>\n

ii) Divisible by 3:
\nA number is divisible by 3 only when sum of the digits in that number is a multiple of 3.
\nSum of the given 5 digits = 2 + 3 + 5 + 6 + 8 = 24.
\nThe 4 digits such that, their sum is a multiple of 3 from the given digits are 2, 3, 5, 8 (or) 2, 5, 6, 8.
\nIn each case, we can permute (arrange) in 4! ways.
\n\u2234 The no. of 4 digited numbers divisible by 3 = 2 . 4!
\n= 2 . 4 . 3 . 2 . 1 = 48.
\n\u2234 The no. of 4 digited numbers divisible by 3 = 48.<\/p>\n

iii) Divisible by 4 :<\/p>\n

\"TS<\/p>\n

A number is divisible by 4.
\nHence, the last two places should be filled with one of the following 28, 32, 36, 52, 56, 68.
\nThus, the last two places can be filled in 6 ways.
\nThe remaining two places can be filled by remaining three digits in \\({ }^3 \\mathrm{P}_2\\) = 3 . 2 = 6 ways.
\nThe number of 4 digited number divisible by 4 = 6 \u00d7 6 = 36.<\/p>\n

iv) Divisible by 5 :<\/p>\n

\"TS<\/p>\n

A number is divisible by 5 when its units place must be filled by 5 from the given integers 2, 3, 5, 6, 8.
\nThis can be done in one way.
\nThe remaining three places can be filled with remaining 4 digits in 4P3 ways = 4 . 3 . 2 = 24 ways.
\n\u2234 The number of 4 digited number divisible by 5 = 1 \u00d7 24 = 24.<\/p>\n

v) Divisible by 25 :<\/p>\n

\"TS<\/p>\n

A number is divisible by 25, when its last two places are filled with 25.
\nThus, the last two places can be filled in one way.
\nThe remaining two places from the remaining 3 digits can be filled in \\({ }^3 \\mathrm{P}_2\\) = 6 ways.
\n\u2234 The number of 4 digited number divisible by 25 = 1 \u00d7 6 = 6.<\/p>\n

Question 29.
\nIf the letters of the word BRING are per-muted in all possible ways and the words thus formed are arranged in the dictionary order, then find the 59th word.
\nSolution:
\nGiven word is BRING.
\n\u2234 The alphabetical order of the letters is B, G, I, N, R.
\nIn the dictionary order, first we write all words beginning with B.
\nClearly the number of words beginning with B are 4! = 24.
\nSimilarly the number of words begin with G are 4! = 24.
\nSince the words begin with B and G sum to 48, the 59th word must start with I.
\nNumber of words given with IB = 3! = 6
\nHence the 59th word must start with IG. Number of words begin with IGB = 2! = 2
\nNumber of words begin with IGN = 2! = 2
\n\u2234 Next word is 59th = IGRBN.<\/p>\n

Question 30.
\nFind the number of 5-letter words that can be formed using the letters of the word MIXTURE which begin with an vowel when repetitions are allowed.
\nSolution:
\nWe have to fill up 5 blanks using the letters of the word MIXTURE having 7 letters among which there are 3 vowels. Fill the first place with one of the vowels (I or U or E) in 3 ways as shown below.<\/p>\n

\"TS<\/p>\n

Each of the remaining 4 places can be filled in 7 ways (since we can use all 7 letters each time).
\nThus the number of 5 letter words is 3 \u00d7 7 \u00d7 7 \u00d7 7 \u00d7 7 = 3 \u00d7 74<\/sup>.<\/p>\n

\"TS<\/p>\n

Question 31.
\nFind the number of 4 – digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 6, that are divisible by
\ni) 2
\nii) 3 when repetition is allowed.
\nSolution:
\ni) Numbers divisible by 2:<\/p>\n

\"TS<\/p>\n

For a number to be divisible by 2.
\nThe units place should be filled with an even digit (2 or 4 or 6).
\nThis can be done in 3 ways.
\nNow, each of the remaining 3 places can be filled in 6 ways.
\n\u2234 The number of 4 digit numbers that are divisible by 2 is 8.<\/p>\n

ii) Numbers divisible by 3 :<\/p>\n

\"TS<\/p>\n

Fill the first three places with the given 6 digits in 63<\/sup> ways.
\nNow, after filling up the first three places with three digits if we fill up the unit’s place in 6 ways we get 6 consecutive positive integers.
\nOut of any six consecutive integers exactly two are divisible by 3.
\nHence, the unit’s place can be filled in two ways.
\n\u2234 The number of 4 digit numbers divisible by 3 is
\n2 . 63<\/sup> = 2 . 216 = 432.<\/p>\n

Question 32.
\nFind the number of 5 – letter words that! can be formed using the letters of the word EXPLAIN that begin and end with a vowel when repetitions are allowed.
\nSolution:<\/p>\n

\"TS<\/p>\n

We can fill the first and last places with vowels each in 3 ways (E, A, I).
\nNow, each of the remaining three places can be filled in 7 ways (when repetitions are allowed).
\n\u2234 The number of 5 letter words which begin and end with vowels is 32<\/sup> . 73<\/sup>
\n= 9 . 343 = 3087.<\/p>\n

\"TS<\/p>\n

Question 33.
\nFind the number of ways of arranging 8 persons around a circular table if two particular persons were to sit together.
\nSolution:
\nTreat the two particular persons as one unit.
\nThen we have 6 + 1 = 7 things.
\nThey can be arranged around a circular table in (7 – 1) ! = 6!
\nNow, the two particular persons can be arranged among themselves in 2! ways.
\n\u2234 The number of required arrangements is 6! – 2! = 6 . 5 . 4 . 3 . 2 . 1 . 2 = 1440.<\/p>\n

Question 34.
\nFind the number of ways of arranging 8 men and 4 women around a circular table. In how many of them
\ni) all the women come together
\nii) no two women come together.
\nSolution:
\nTotal number of persons = 12 [8M + 4W]
\n\u2234 The number of ways of arranging these 12 persons around a circular table = (12 – 1)1 = 11!
\ni) All the women come together :
\nTreat the four women as one unit then we have 8 men + 1 unit of women then we have 9 entries.
\nThey can be arranged around a circular table in (9 – 1) ! = 8! ways.
\nNow, the four women among themselves can be arranged in 4! ways.
\nThe required number of arrangements is 8! . 4!.<\/p>\n

ii) No two women come together :<\/p>\n

\"TS<\/p>\n

First we arrange 8 men around the circular table in (8 – 1) ! = 7 ! ways.
\nThere are 8 places in between them.
\nNow, we can arrange the four women in these 8 places in \\({ }^8 \\mathrm{P}_4\\) ways.
\nThe required number of arrangements is 7 ! . \\({ }^8 \\mathrm{P}_4\\).<\/p>\n

Question 35.
\nFind the number of ways of seating 5 Indians, 4 Americans and 3 Russians at a round table so that
\ni) all Indians sit together
\nii) no two Russians sit together
\niii) persons of same nationality sit together.
\nSolution:
\ni) All Indians sit together :
\nTreat the five Indians as one unit.
\nThen we have 4 Americans + 3 Russians + 1 unit of Indians = 8 entries.
\nThey can be arranged at a round table in (8 – 1) ! = 7 ! ways.
\nNow, the five Indians among themselves can be arranged in 5! ways.
\n\u2234 The required number of arrangements is 7! – 5!.<\/p>\n

ii) Now two Russians sit together :
\nFirst we arrange the 5 Indians + 4 Americans around the table in (9 – 1)! = 8! ways.
\nNow, there are 9 gaps in between 9 persons.
\nThe 3 Russians can be arranged in the 9 gaps in \\({ }^9 P_3\\).
\n\u2234 The required number of arrangements is 8! . \\({ }^9 P_3\\).<\/p>\n

Persons of same nationality sit together:
\nTreat the 5 Indians as one unit, the four Americans as one unit and the 3 Russians as one unit.
\nThese 3 units can be arranged at round table in (3 – 1)1 = 2! ways.
\nNow, the 5 Indians among themselves can be arranged in 5! ways.
\nThe four Americans among themselves can be arranged in 4! ways.
\nThe 3 Russians among themselves can be arranged in 3! ways.
\nThe required number of arrangements = 2! . 5! . 3! . 4!.<\/p>\n

\"TS<\/p>\n

Question 36.
\nFind the number of ways of arranging the letters of the word SPECIFIC. In how many of them
\ni) the two ‘C’, ‘S’ come together,
\nii) the two I’s do not come together.
\nSolution:
\nThe given word has eight letters in which there are 2 I’s are alike and 2 C’s are alike and rest are different.
\nThey can be arranged in \\(\\frac{8 !}{2 ! 2 !}\\) ways.
\n\\(\\frac{8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1}{2 \\cdot 2}\\) = 10080.<\/p>\n

i) the two C\u2019s come together:
\nTreat the 2 C\u2019s as one unit.
\nThen we have 6 + 1 = 7 letters.
\nIn which 2 I\u2019s are alike.
\nNow, they can be arranged in \\(\\frac{7 !}{2 !}\\) ways.
\nThe 2 C’s can be arranged among themselves in \\(\\frac{2 !}{2 !}\\) ways.
\n\u2234 The number of required arrangements = \\(\\frac{7 !}{2 !} \\cdot \\frac{2 !}{2 !}\\) = 2520.<\/p>\n

ii) The 2I’s do not come together:
\nSince, the 2 I’s do not come togeter, then first arrange the remaining six letters in \\(\\frac{6 !}{2 !}\\) ways.
\nAmong these 6 letters we find 7 gaps.
\nNow, the 2 I\u2019s can be arranged in these 7 gaps in \\(\\frac{{ }^7 \\mathrm{P}_2}{2 !}\\) ways.
\n\u2234 The number of required arrangements = \\(\\frac{6 !}{2 !} \\cdot \\frac{{ }^7 \\mathrm{P}_2}{2 !}\\)
\n= \\(\\frac{6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1}{2} \\cdot \\frac{7 \\cdot 6}{2}\\) = 7560<\/p>\n

Question 37.
\nFind the number of 5- digit numbers that can be formed using the digits 1, 1, 2, 2, 3. How many of them are even?
\nSolution:
\nIn the given 5 digits there are two 1\u2019s and two 2\u2019s.
\nThey can be arranged in \\(\\frac{5 !}{2 ! 2 !}\\) ways
\n= \\(\\frac{5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1}{2 \\cdot 2}\\) = 30
\nNow, the remaining 4 places can be arranged using the remaining digits 1, 1, 2, 3 in \\(\\frac{4 !}{2 !}\\) ways = 12.
\nThe number of 5 digit even numbers that can be formed using the digits 1, 1, 2, 2, 3 is 12.<\/p>\n

Question 38.
\nFind the number of ways of arranging the letters of the words if
\ni) SINGING
\nii) PERMUTATION
\niii) COMBINATION
\nSolution:
\ni) SINGING: Given word is ‘SINGING’.
\nThe word SINGING contains 7 letters in which there are 2 I’s are alike, 2 N’s are alike and 2G’s are alike, and rest are different.
\n\u2234 The number of required arrangements = \\(\\frac{7 !}{2 ! 2 ! 2 !}\\).<\/p>\n

ii) PERMUTATION:
\nGiven word is \u2018PERMUTATION\u2019.
\nThe word \u2018PERMUTATION\u2019 contains 11 letters in which there are 2 T\u2019s are alike and rest are different.
\n\u2234 The number of required arrangements = \\(\\frac{11 !}{2 !}\\)<\/p>\n

iii) COMBINATION:
\nGiven word is COMBINATiON.
\nThe word \u2018COMBiNATION\u2019 contains 11 letters in which there are 2 O\u2019s are alike, 2 I\u2019s are alike, 2 N\u2019s are alike and rest are different.
\n\u2234 The number of required arrangements = \\(\\frac{11 !}{2 ! 2 ! 2 !}\\).<\/p>\n

\"TS<\/p>\n

Question 39.
\nIf the letters of the word AJANTA are permuted in all possible ways and the words thus formed are arranged in dictionary order, find the ranks of the words
\n(i) AJANTA
\n(ii) JANATA.
\nSolution:
\nThe alphabetical order of the letters of the given word is AAAJNT.<\/p>\n

\"TS<\/p>\n

The no. of words begin with AA is 4! = 24
\nThe no. of words begin with AJAA is 2! = 2
\nThe no. of words begin with AJANA is 1! = 1
\nThe next word is AJANTA = 1
\nRank of the word, AJANTA = 24 + 2 + 2 + 1 + 1 = 28.<\/p>\n

ii) The alphabetical order of the letters of the given word is AAAJNT.<\/p>\n

\"TS<\/p>\n

The no. of words begin with A is \\(\\frac{5 !}{2 !}\\) = 60
\nThe no. of words begin with JAA is 3! = 6
\nThe no. of words begin with JANAA is 1! = 1
\nThe next word \u00a1s JANATA = I
\n\u2234 Rank of the word, JANATA = 60 + 6 + 1 + 1 = 68.<\/p>\n

Question 40.
\nFind the number of ways of forming a committee of 4 members out of 6 boys and 4 girls such that there is atleast one girl in the committee.
\nSolution:<\/p>\n

\"TS<\/p>\n

Since, the committee must contain at least one girl, the numbers of committee will be as follows:
\ni) 3 boys and one girl :
\nThe number of ways of selecting committee = \\({ }^6 \\mathrm{C}_3 \\cdot{ }^4 \\mathrm{C}_1=\\frac{6 \\cdot 5 \\cdot 4}{3 \\cdot 2 \\cdot 1} \\cdot 4\\) = 80.<\/p>\n

ii) 2 boys and 2 girls :
\nThe number of ways of selecting committee = \\({ }^6 \\mathrm{C}_2 \\cdot{ }^4 \\mathrm{C}_2=\\frac{6 \\cdot 5}{2 \\cdot 1} \\cdot \\frac{4 \\cdot 3}{2 \\cdot 1}\\) = 90.<\/p>\n

iii) One boy and 3 girls :
\nThe number of ways of selecting committee = \\({ }^6 \\mathrm{C}_1 \\cdot{ }^4 \\mathrm{C}_3\\) = 6 . 4 = 24<\/p>\n

iv) 0 boys and 4 girls: The number of ways of selecting committee = \\({ }^6 \\mathrm{C}_0 \\cdot{ }^4 \\mathrm{C}_4\\) = 1
\n\u2234 The required number of ways of selecting a committee is = 80 + 90 + 24 + 1 = 195.<\/p>\n

\"TS<\/p>\n

Question 41.
\nFind the number of ways of selecting 11 member cricket team from 7 batsmen, 6 bowlers and 2 wicket keepers so that the team contains 2 wicket keepers and atleast 4 bowlers. [March \u201814]
\nSolution:<\/p>\n

\"TS<\/p>\n

The number of ways of selecting the required cricket team = 210 + 252 + 70 + 315 + 210 + 35 = 1092.<\/p>\n

Question 42.
\nIf 2 \u2264 r \u2264 n, then show that \\({ }^n C_{(r-1)}+2 \\cdot{ }^n C_{(r-1)}+{ }^n C_r={ }^{(n+2)} C_r\\).
\nSolution:
\nL.H.S:
\n\\({ }^n C_{(r-1)}+2 \\cdot{ }^n C_{(r-1)}+{ }^n C_r\\)
\n= \\({ }^n C_{r-2}+{ }^n C_{r-1}+{ }^n C_{r-1}+{ }^n C_r\\)
\n= \\({ }^{(n+1)} C_{r-1}+{ }^{(n-1)} C_r={ }^{(n+2)} C_r\\) = R.H.S<\/p>\n

Question 43.
\n14 persons are seated at a round table. Find the number of ways of selecting two persons out of them who are not seated
\nadjacent to each other. [AP – May 2015]
\nSolution:<\/p>\n

\"TS<\/p>\n

14 persons at the round table.
\nSelecting 2 persons from 14 persons to be seated on a circular table = \\({ }^{14} C_2=\\frac{14 \\cdot 13}{1 \\cdot 2}\\) = 91 ways.
\nThe number of ways of selecting 2 adjacent persons 14 (a1<\/sub>a2<\/sub>, a2<\/sub>a3<\/sub>, a3<\/sub>a4<\/sub> ………………….. a14<\/sub>a1<\/sub>)
\n\u2234 The number of required selections = 91 – 14 = 77.<\/p>\n

\"TS<\/p>\n

Question 44.
\nFind the number of ways of selecting a committee of 6 members out of 10 members always including a specified number.
\nSolution:
\nSince, a specified member always includes in a committee.
\nRemaining 5 members can be selected from 9 members in \\({ }^9 \\mathrm{C}_5\\) ways.
\n\u2234 Required number of ways of selecting a committee C = \\({ }^9 \\mathrm{C}_5=\\frac{9 \\cdot 8 \\cdot 7 \\cdot 6 \\cdot 5}{5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1}\\) = 126.<\/p>\n

Question 45.
\nIf a set A has 12 elements, find the num\u00acber of subsets of A having
\ni) 4 elements
\nii) Atleast 3 elements
\niii) Atmost 3 elements.
\nSolution:
\nNumber of elements in set A = 12
\ni) Number of subsets of A with exactly four elements = \\({ }^{{ }^n} \\mathrm{C}_{\\mathrm{r}}={ }^{12} \\mathrm{C}_4\\)
\n= \\(\\frac{12 \\cdot 11 \\cdot 10 \\cdot 9}{4 \\cdot 3 \\cdot 2 \\cdot 1}\\) = 495.<\/p>\n

ii) Number of subsets of A with exactly zero elements is \\({ }^{12} \\mathrm{C}_0\\).
\nNumber of subsets of A with exactly one elements is \\({ }^{12} \\mathrm{C}_1\\).
\nNumber of subsets of A with exactly two elements is \\({ }^{12} \\mathrm{C}_2\\).
\nTotal number of subsets of A formed = 2n<\/sup> = 212<\/sup>.
\nNumber 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> – [latex]{ }^{12} \\mathrm{C}_0+{ }^{12} \\mathrm{C}_1+{ }^{12} \\mathrm{C}_2[\/latex]
\n= 4096 – 1 – 12 – \\(\\frac{12 \\cdot 11}{2 \\cdot 1}\\)
\n= 4096 – 1 – 12 – 66
\n= 4096 – 79 = 4017.<\/p>\n

iii) The required subsets contain atmost 3 elements i.e., it may contain 0 or 1 or 2 or 3 elements.
\nNumber of subsets of A with exactly one element is \\({ }^{12} \\mathrm{C}_1\\).
\nNumber of subsets of A with exactly two elements is \\({ }^{12} \\mathrm{C}_2\\).
\nNumber of subsets of A with exactly three elements is \\({ }^{12} \\mathrm{C}_3\\).
\n:. Number of subsets of A with atmost 3 elements = \\({ }^{12} \\mathrm{C}_0+{ }^{12} \\mathrm{C}_1+{ }^{12} \\mathrm{C}_2+{ }^{12} \\mathrm{C}_3\\)
\n= 1 + 12 + 66 + \\(\\frac{12 \\cdot 11 \\cdot 10}{3 \\cdot 2 \\cdot 1}\\)
\n= 79 + 220 = 299<\/p>\n

Question 46.
\nIn 5 vowels and 6 consonants are given, then how many 6 letter words can be formed with 3 vowels and 3 consonants.
\nSolution:
\nGiven 5 vowels and 6 consonants.
\n6 letter word is formed with 3 vowels and 3 consonants.
\nNumber of ways of selecting 3 vowels from 5 vowels is \\({ }^5 \\mathrm{C}_3\\).
\nNumber of ways of selecting 3 consonants is from 6 consonants is \\({ }^6 \\mathrm{C}_3\\).
\n\u2234 Total number of ways of selecting = \\({ }^5 \\mathrm{C}_3 \\times{ }^6 \\mathrm{C}_3\\)
\nThese letters can be arranged themselves in 6! ways.
\n\u2234 Number of 6 letter words formed = \\({ }^5 \\mathrm{C}_3 \\times{ }^6 \\mathrm{C}_3\\) \u00d7 6!.<\/p>\n

\"TS<\/p>\n

Question 47.
\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 atleast one from each section.
\nSolution:
\nA question paper contains 3 sections A, B, C containing 3, 4, 5 questions respectively.
\nNumber of ways of selecting 6 questions out of these 12 questions = \\({ }^{12} \\mathrm{C}_6\\)
\nNumber of ways of selecting 6 questions from sections B and C (i.e., from 9 questions) = \\({ }^9 \\mathrm{C}_6\\)
\nNumber of ways of selecting 6 questions from sections A and C (i.e., from 8 questions) = \\({ }^8 \\mathrm{C}_6\\)
\nNumber of ways of selecting 6 questions from sections A and B (i.e., 7 questions) = \\({ }^7 \\mathrm{C}_6\\)
\n\u2234 Number of ways of selecting 6 questions choosing atleast one from each section = \\({ }^{12} \\mathrm{C}_6-{ }^7 \\mathrm{C}_6-{ }^8 \\mathrm{C}_6-{ }^9 \\mathrm{C}_6\\) = 805.<\/p>\n

Question 48.
\nFind the sum of all 4-digit numbers that can be formed using the digits 0, 2, 4, 7, 8 without repetition.
\nSolution:
\nThe number of 4 digited numbers formed by the using the digits 0, 2, 4, 7 8 = \\(\\) (\u2235 0 is present).
\n= 5 \u00d7 4 \u00d7 3 \u00d7 2 – 4 \u00d7 3 \u00d7 2
\n= 120 – 24 = 96
\nThe sum of r-digited numbers formed with n distinct non-zero digits (when \u20190′ is present).
\n= \\({ }^{(n-1)} P_{(r-1)}\\) \u00d7 (Sum of \u2019n\u2019 digits) \u00d7 (111 …………….. 1)r times<\/sub> – \\({ }^{(n-2)} \\mathrm{P}_{(\\mathrm{r}-2)}\\) \u00d7 (Sum of n digits) \u00d7 (111 …………. 1)r – 1 times<\/sub>
\n= \\((5-1)_{P_{(4-1)}}\\) \u00d7 (0 + 2 + 4 + 7 + 8) \u00d7 1111 – \\((5-2)_{P_{(4-2)}}\\) \u00d7 (0 + 2 + 4 + 7 + 8) \u00d7 111
\n= \\({ }^4 \\mathrm{P}_3\\) \u00d7 21 \u00d7 1111 – \\({ }^3 \\mathrm{P}_2\\) \u00d7 21 \u00d7 111
\n= 24 \u00d7 21 \u00d7 1111 – 6 \u00d7 21 \u00d7 111 = 5,45,958.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Permutations and Combinations Important Questions Short Answer Type to help strengthen their preparations for exams. TS Inter Second Year Maths 2A Permutations and Combinations Important Questions Short Answer Type Question 1. 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