{"id":17472,"date":"2022-05-24T23:00:58","date_gmt":"2022-05-24T17:30:58","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=17472"},"modified":"2022-05-16T15:44:24","modified_gmt":"2022-05-16T10:14:24","slug":"mcq-questions-for-class-11-maths-chapter-7","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-11-maths-chapter-7\/","title":{"rendered":"MCQ Questions for Class 11 Maths Chapter 7 Permutations and Combinations with Answers"},"content":{"rendered":"

Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 7 Permutations and Combinations with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers<\/a> during preparation and score maximum marks in the exam. Students can download the Permutations and Combinations Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 7 Permutations and Combinations Objective Questions.<\/p>\n

Permutations and Combinations Class 11 MCQs Questions with Answers<\/h2>\n

Students are advised to solve the Permutations and Combinations Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Permutations and Combinations Class 11 with answers will boost your confidence thereby helping you score well in the exam.<\/p>\n

Explore numerous MCQ Questions of Permutations and Combinations Class 11 with answers provided with detailed solutions by looking below.<\/p>\n

Question 1.
\nIt is required to seat 5 men and 4 women in a row so that the women occupy the even places. The number of ways such arrangements are possible are
\n(a) 8820
\n(b) 2880
\n(c) 2088
\n(d) 2808<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 2880
\nTotal number of persons are 9 in which there are 5 men and 4 women
\nSo total number of place = 9
\nNow women seat in even place
\nSo total number of arrangement = 4! (_W_W_W_W_) (W-Woman)
\nMen sit in odd place
\nSo total number of arrangement = 5! (MWMWMWMWM) (M-Man)
\nNow Total number of arrangement = 5! \u00d7 4! = 120 \u00d7 24 = 2880<\/p>\n<\/details>\n

\n

Question 2.
\nSix boys and six girls sit along a line alternately in x ways and along a circle (again alternatively in y ways), then
\n(a) x = y
\n(b) y = 12x
\n(c) x = 10y
\n(d) x = 12y<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) x = 12y
\nGiven, six boys and six girls sit along a line alternately in x ways and along a circle
\n(again alternatively in y ways).
\nNow, x = 6! \u00d7 6! + 6! \u00d7 6!
\n\u21d2 x = 2 \u00d7 (6!)2
\nand y = 5! \u00d7 6!
\nNow, x\/y = {2 \u00d7 (6!)2}\/(5! \u00d7 6!)
\n\u21d2 x\/y = {2 \u00d7 6! \u00d7 6! }\/(5! \u00d7 6!)
\n\u21d2 x\/y = {2 \u00d7 6!}\/5!
\n\u21d2 x\/y = {2 \u00d7 6 \u00d7 5!}\/5!
\n\u21d2 x\/y = 12
\n\u21d2 x = 12y<\/p>\n<\/details>\n

\n

Question 3.
\nHow many 3-letter words with or without meaning, can be formed out of the letters of the word, LOGARITHMS, if repetition of letters is not allowed
\n(a) 720
\n(b) 420
\n(c) none of these
\n(d) 5040<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 720
\nThe word LOGARITHMS has 10 different letters.
\nHence, the number of 3-letter words(with or without meaning) formed by using these letters
\n= 10<\/sup>P3<\/sub>
\n= 10 \u00d7 9 \u00d7 8
\n= 720<\/p>\n<\/details>\n

\n

Question 4.
\nA committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of at least 3 girls
\n(a) 588
\n(b) 885
\n(c) 858
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 588
\nGiven number of boys = 9
\nNumber of girls = 4
\nNow, A committee of 7 has to be formed from 9 boys and 4 girls.
\nNow, the committee consists of atleast 3 girls:
\n4<\/sup>C3<\/sub> \u00d7 9<\/sup>C4<\/sub> + 4<\/sup>C4<\/sub> \u00d7 9<\/sup>C3<\/sub>
\n= [{4! \/ (3! \u00d7 1!)} \u00d7 {9! \/ (4! \u00d7 5!)}] + 9C3<\/sub>
\n= [{(4 \u00d7 3!) \/3!} \u00d7 {(9 \u00d7 8 \u00d7 7 \u00d7 6 \u00d7 5!) \/ (4! \u00d7 5!)}] + 9! \/(3! \u00d7 6!)
\n= [4 \u00d7 {(9 \u00d7 8 \u00d7 7 \u00d7 6) \/ 4!}] + (9\u00d78\u00d77\u00d76!)\/(3! \u00d7 6!)
\n= [{4 \u00d7 (9 \u00d7 8 \u00d7 7 \u00d7 6)} \/ (4 \u00d7 3 \u00d7 2 \u00d7 1)] + (9 \u00d7 8 \u00d7 7)\/3!
\n= (9 \u00d7 8 \u00d7 7) + (9 \u00d7 8 \u00d7 7)\/(3 \u00d7 2 \u00d7 1)
\n= 504 + (504\/6)
\n= 504 + 84
\n= 588<\/p>\n<\/details>\n

\n

Question 5.
\nIn how many ways can 12 people be divided into 3 groups where 4 persons must be there in each group?
\n(a) none of these
\n(b) 12!\/(4!)\u00b3
\n(c) Insufficient data
\n(d) 12!\/{3! \u00d7 (4!)\u00b3}<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 12!\/{3! \u00d7 (4!)\u00b3}
\nNumber of ways in which
\nm \u00d7 n”>
\nm \u00d7 n distinct things can be divided equally into n
\nn”> groups
\n= (mn)!\/{n! \u00d7 (m!)n }
\nGiven, 12(3 \u00d7 4) people needs to be divided into 3 groups where 4 persons must be there in each group.
\nSo, the required number of ways = (12)!\/{3! \u00d7 (4!)n}<\/p>\n<\/details>\n

\n

Question 6.
\nHow many factors are 25<\/sup> \u00d7 36<\/sup> \u00d7 5\u00b2 are perfect squares
\n(a) 24
\n(b) 12
\n(c) 16
\n(d) 22<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 24
\nAny factors of 25<\/sup> \u00d7 36<\/sup> \u00d7 5\u00b2 which is a perfect square will be of the form 2a<\/sup> \u00d7 3b<\/sup> \u00d7 5c<\/sup>
\nwhere a can be 0 or 2 or 4, So there are 3 ways
\nb can be 0 or 2 or 4 or 6, So there are 4 ways
\na can be 0 or 2, So there are 2 ways
\nSo, the required number of factors = 3 \u00d7 4 \u00d7 2 = 24<\/p>\n<\/details>\n

\n

Question 7.
\nIf \u207fC15<\/sub> = \u207fC6<\/sub> then the value of \u207fC21<\/sub> is
\n(a) 0
\n(b) 1
\n(c) 21
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 1
\nWe know that
\nif \u207fCr1<\/sub> = \u207fCr2<\/sub>
\n\u21d2 n = r1<\/sub> + r2<\/sub>
\nGiven, \u207fC15<\/sub> = \u207fC6<\/sub>
\n\u21d2 n = 15 + 6
\n\u21d2 n = 21
\nNow, 21<\/sup>C21<\/sub> = 1<\/p>\n<\/details>\n

\n

Question 8.
\nIf n+1<\/sup>C3<\/sub> = 2 \u207fC2<\/sub>, then the value of n is
\n(a) 3
\n(b) 4
\n(c) 5
\n(d) 6<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 6
\nGiven, n+1<\/sup>C3<\/sub> = 2 \u207fC2<\/sub>
\n\u21d2 [(n + 1)!\/{(n + 1 – 3) \u00d7 3!}] = 2n!\/{(n – 2) \u00d7 2!}
\n\u21d2 [{n \u00d7 n!}\/{(n – 2) \u00d7 3!}] = 2n!\/{(n – 2) \u00d7 2}
\n\u21d2 n\/3! = 1
\n\u21d2 n\/6 = 1
\n\u21d2 n = 6<\/p>\n<\/details>\n

\n

Question 9.
\nThere are 15 points in a plane, no two of which are in a straight line except 4, all of which are in a straight line. The number of triangle that can be formed by using these 15 points is
\n(a) 15<\/sup>C3<\/sub>
\n(b) 490
\n(c) 451
\n(d) 415<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 451
\nThe required number of triangle = 15<\/sup>C3<\/sub> – 4<\/sup>C3<\/sub> = 455 – 4 = 451<\/p>\n<\/details>\n

\n

Question 10.
\nIn how many ways in which 8 students can be sated in a circle is
\n(a) 40302
\n(b) 40320
\n(c) 5040
\n(d) 50040<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 5040
\nThe number of ways in which 8 students can be sated in a circle = ( 8 – 1)!
\n= 7!
\n= 5040<\/p>\n<\/details>\n

\n

Question 11.
\nLet R = {a, b, c, d} and S = {1, 2, 3}, then the number of functions f, from R to S, which are onto is
\n(a) 80
\n(b) 16
\n(c) 24
\n(d) 36<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 36
\nTotal number of functions = 34<\/sup> = 81
\nAll the four elements can be mapped to exactly one element in 3 ways, and exactly 3
\nelements in 3(24<\/sup> – 2) = 3(16 – 2) = 3 \u00d7 14 = 42
\nThus the number of onto functions = 81 – 42 -3 = 81 – 45 = 36<\/p>\n<\/details>\n

\n

Question 12.
\nIf (1 + x)\u207f = C0<\/sub> + C1<\/sub> x + C2<\/sub> x\u00b2 + …………..+ Cn<\/sub> x\u207f, then the value of C0<\/sub>\u00b2 + C1<\/sub>\u00b2 + C2<\/sub>\u00b2 + …………..+ Cn<\/sub>\u207f = \u00b2\u207fCn<\/sub> is
\n(a) (2n)!\/(n!)
\n(b) (2n)!\/(n! \u00d7 n!)
\n(c) (2n)!\/(n! \u00d7 n!)2
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) (2n)!\/(n! \u00d7 n!)
\nGiven, (1 + x)\u207f = C0<\/sub> + C1<\/sub> x + C2<\/sub> x\u00b2 + ………….. + Cn<\/sub> x\u207f ………. 1
\nand (1 + x)\u207f = C0<\/sub> x\u207f + C1<\/sub> xn-1<\/sup> + C2<\/sub> xn-2<\/sup> + ………….. Cr<\/sub> xn-r<\/sup> + ………. + Cn-1<\/sub> x + Cn<\/sub> ……….. 2
\nMultiply 1 and 2, we get
\n(1 + x)\u00b2\u207f = (C0<\/sub> + C1<\/sub> x + C2<\/sub> x\u00b2 + …………..+ Cn<\/sub> x\u207f) \u00d7 (C0<\/sub> x\u207f + C1<\/sub> xn-1<\/sup> + C2<\/sub> xn-2<\/sup> + ………….. Cr<\/sub> xn-r<\/sup> + ………. + Cn-1<\/sub> x + Cn<\/sub>)
\nNow, equating the coefficient of xn on both side, we get
\nC0<\/sub>\u00b2 + C1<\/sub>\u00b2 + C2<\/sub>\u00b2 + …………..+ Cn<\/sub>\u207f = \u00b2\u207fCn<\/sub> = (2n)!\/(n! \u00d7 n!)<\/p>\n<\/details>\n

\n

Question 13.
\nThe total number of 9 digit numbers of different digits is
\n(a) 99!
\n(b) 9!
\n(c) 8 \u00d7 9!
\n(d) 9 \u00d7 9!<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 9 \u00d7 9!
\nGiven digit in the number = 9
\n1st place can be filled = 9 ways = 9 (from 1-9 any number can be placed at first position)
\n2nd place can be filled = 9 ways (from 0-9 any number can be placed except the number which is placed at the first position)
\n3rd place can be filled = 8 ways
\n4th place can be filled = 7 ways
\n5th place can be filled = 6 ways
\n6th place can be filled = 5 ways
\n7th place can be filled = 4 ways
\n8th place can be filled = 3 ways
\n9th place can be filled = 2 ways
\nSo total number of ways = 9 \u00d7 9 \u00d7 8 \u00d7 7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2
\n= 9 \u00d7 9!<\/p>\n<\/details>\n

\n

Question 14.
\nThe number of ways in which 6 men add 5 women can dine at a round table, if no two women are to sit together, is given by
\n(a) 30
\n(b) 5 ! \u00d7 5 !
\n(c) 5 ! \u00d7 4 !
\n(d) 7 ! \u00d7 5 !<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 5 ! \u00d7 5 !
\nAgain, 6 girls can be arranged among themselves in 5! ways in a circle.
\nSo, the number of arrangements where boys and girls sit attentively in a circle = 5! \u00d7 5!<\/p>\n<\/details>\n

\n

Question 15.
\nThere are 15 points in a plane, no two of which are in a straight line except 4, all of which are in a straight line. The number of triangle that can be formed by using these 15 points is
\n(a) 15<\/sup>C3<\/sub>
\n(b) 490
\n(c) 451
\n(d) 415<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 451
\nThe required number of triangle = 15<\/sup>C3<\/sub> – 4<\/sup>C3<\/sub> = 455 – 4 = 451<\/p>\n<\/details>\n

\n

Question 16.
\nThe number of 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated are
\n(a) 110
\n(b) 120
\n(c) 130
\n(d) 140<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 120
\nA number is divisible by 10 if the unit digit of the number is 0.
\nGiven digits are 0, 1, 3, 5, 7, 9
\nNow we fix digit 0 at unit place of the number.
\nRemaining 5 digits can be arranged in 5! ways
\nSo, total 6-digit numbers which are divisible by 10 = 5! = 120<\/p>\n<\/details>\n

\n

Question 17.
\n6 men and 4 women are to be seated in a row so that no two women sit together. The number of ways they can be seated is
\n(a) 604800
\n(b) 17280
\n(c) 120960
\n(d) 518400<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 604800
\n6 men can be sit as
\n\u00d7 M \u00d7 M \u00d7 M \u00d7 M \u00d7 M \u00d7 M \u00d7
\nNow, there are 7 spaces and 4 women can be sit as 7<\/sup>P4<\/sub> = 7<\/sup>P3<\/sub> = 7!\/3! = (7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3!)\/3!
\n= 7 \u00d7 6 \u00d7 5 \u00d7 4 = 840
\nNow, total number of arrangement = 6! \u00d7 840
\n= 720 \u00d7 840
\n= 604800<\/p>\n<\/details>\n

\n

Question 18.
\nA committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of exactly 3 girls
\n(a) 540
\n(b) 405
\n(c) 504
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 504
\nGiven number of boys = 9
\nNumber of girls = 4
\nNow, A committee of 7 has to be formed from 9 boys and 4 girls.
\nNow, If in committee consist of exactly 3 girls:
\n4<\/sup>C3<\/sub> \u00d7 9<\/sup>C4<\/sub>
\n= {4! \/ (3! \u00d7 1!)} \u00d7 {9! \/ (4! \u00d7 5!)}
\n= {(4\u00d73!) \/3!} \u00d7 {(9 \u00d7 8 \u00d7 7 \u00d7 6 \u00d7 5!) \/ (4! \u00d7 5!)}
\n= 4 \u00d7 {(9 \u00d7 8 \u00d7 7 \u00d7 6) \/ 4!}
\n= {4 \u00d7 (9 \u00d7 8 \u00d7 7 \u00d7 6)} \/ (4 \u00d7 3 \u00d7 2 \u00d7 1)
\n= 9 \u00d7 8 \u00d7 7
\n= 504<\/p>\n<\/details>\n

\n

Question 19.
\nHow many factors are 25<\/sup> \u00d7 36<\/sup> \u00d7 5\u00b2 are perfect squares
\n(a) 24
\n(b) 12
\n(c) 16
\n(d) 22<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 24
\nAny factors of 25<\/sup> \u00d7 36<\/sup> \u00d7 5\u00b2 which is a perfect square will be of the form 2a<\/sup> \u00d7 3b<\/sup> \u00d7 5c<\/sup>
\nwhere a can be 0 or 2 or 4, So there are 3 ways
\nb can be 0 or 2 or 4 or 6, So there are 4 ways
\na can be 0 or 2, So there are 2 ways
\nSo, the required number of factors = 3 \u00d7 4 \u00d7 2 = 24<\/p>\n<\/details>\n

\n

Question 20.
\nThe value of 2 \u00d7 P(n, n-2) is
\n(a) n
\n(b) 2n
\n(c) n!
\n(d) 2n!<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) n!
\nGiven, 2 \u00d7 P(n, n – 2)
\n= 2 \u00d7 {n!\/(n – (n – 2))}
\n= 2 \u00d7 {n!\/(n – n + 2)}
\n= 2 \u00d7 (n!\/2)
\n= n!
\nSo, 2 \u00d7 P(n, n – 2) = n!<\/p>\n<\/details>\n

\n

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Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 7 Permutations and Combinations with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students …<\/p>\n

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