Tiêu đề 7-permutations--combinations.pdf ✅

Permutations Combinations 1. (A) Using de-arrangement theorem, Number of ways to place 6 letters in 6 envelopes such that all are placed in wrong envelopes. = 6 [1 − 1 ⌊1 + 1 ⌊2 − 1 ⌊3 + ⋯ . + 1 ⌊6 ] = 360 − 120 + 30 − 6 + 1 = 265 2. (A) Required sum of number = [Sum of four digit numbers using 0,1,2,3,4, allowing 0 in first place] −[ Sum of three digit numbers using 1,2,3,4)]. = 5! 5 [0 + 1 + 2 + 3 + 4][1 + 10 + 102 + 103 ] − 4! 4 (1 + 2 + 3 + 4)(1 + 10 + 102 ) = 24 × 10 × 1111 − 6 × 10 × 111 = 259980 3. (C) Refer to Solution Illustration-38 4. (B) Refer to Solution Illustration-41 5. (B) It is possible in two mutually exclusive cases Case 1: 2 children get none, one child gets three and all remaining 7 children get one each. Case 2: 2 children get none, 2 children get 2 each and all remaining 6 children get one each. Using formula given in section (6.3), we get : Case I : Number of ways = ( 10! 3! 7! 1 2! ) (10!) Case II : Number of ways = ( 10! (2!) 2 ⋅ 6! 1 2! 1 2! ) 10! Thus total ways = (10!) 2 ( 1 3! 7! 2! + 1 (2!) 46! ) 6. (B) Refer to Solution Illustrating the concepts (ii) : Page-46 7. (A) The 4 men can be seated at the circular table such that there is a vacant seat between every pair of men in (4 − 1)! = 3 ! ways. Now, 4 vacant seats can be occupied by 4 women in 4 ! ways. Hence, the required number of seating arrangements = 3! × 4! = 144. 8. (B) Refer to Solution Illustration-48 9. (A) Refer to Solution Illustration-48 10. (B) A is not included, then others can be selected in 2 k − 1 ways. T is not included, then others can be selected in 2 k − 1 ways. M is not included, then others can be selected in 2 k − 1 ways. 3(2 k − 1) = 93 ⇒ 2 k = 32 ⇒ k = 5 11. (D) Refer to Solution OWS/Question-39 12. (B) Total no. of words = 9! 2! No. of words having "HIN" together = 7 ! No. of words having "DUS" together = 7!/2 ! No. of words having "TAN" together = 7 ! No. of words having "HIN" and "DUS" together = 5 ! No. of words having "HIN" and TAN" together = 5! No. of words having "DUS" and "TAN" together = 5 ! No. of words having HIN, DUS and TAN together = 3 ! No. of words required 9! 2! − (7! + 7! 2! + 7!) + (5! + 5! + 5!) − 3! = 169194 13. (B) Refer to Solution INE-E/Question-10 14. (C) Refer to Solution OWS/Question-13 15. (C) Refer to Solution OWS/Question-25 16. (A) For each of the first (n − 1) elements a1, a2, ... , an−1 we have two choices: either ai(1 ≤ i ≤ n − 1) lies in the subset or ai doesn't lie in the subset. For the last element we have just one choice. If even number of elements have already been selected, we do not include an in the subset, otherwise (when odd number of elements have been selected), we include it in the subset. Thus, the number of subsets of A = {a1, a2, ... , an } which contain even number of elements is equal to 2 n−1 . 17. (A) A number between 100 and 1000 has three digits. So, we have to form all possible 3-digit numbers with distinct digits. We cannot have 0 at the hundred's place. So, the
hundred's place can be filled with any of the 9 digits 1,2,3, ... ,9. So, there are 9 ways of filling the hundred's place. Now, 9 digits are left including 0 . So, ten's place can be filled with any of the remaining 9 digits in 9 ways. Now, the unit's place can be filled with in any of the remaining 8 digits. So, there are 8 ways of filling the unit's place. Hence, the total number of required numbers = 9 × 9 × 8 = 648. 18. (B) This can be done in four mutually exclusive ways as follows : Row R R Row R R Row R Number of ways I. 1 3 2 ( 2C1 )( 4C3 )( 2C2 ) = 8 II. 1 4 1 ( 2C1 )( 4C4 )( 2C1 ) = 4 III. 2 2 2 ( 2C2 )( 4C2 )( 2C2 ) = 6 IV. 2 3 1 ( 2C2 )( 4C3 )( 2C1 ) = 8 Alternate Sol : There are only two ways in which a row will not contain any X ∴ Total number of ways = 8C2 − 2 = 26 19. (A) In dictionary the words at each stage are arranged in alphabetical order. Starting with the letter A, and arranging the other four letters GAIN, we obtain 4! = 24 words. Thus, there are 24 words which start with A. These are the first 24 words. Then, starting with G, and arranging the other four letters A, A, I, N in different ways, we obtain 4! 2! = 24 2 = 12 words. Thus, there are 12 words, which start with G. Now, we start with I. The remaining 4 letters A, G, A, N can be arranged in 4! 2! = 12 ways. So, there are 12 words, which start with I. Thus, we have so far constructed 48 words. The 49th word is NAAGI and hence the 50th word is NAAIG. 20. (C) 7−xPx−3 is defined when x − 3 ≤ 7 − x ⇒ x ≤ 5 and x − 3 ≥ 0 ⇒ x ≥ 3. ⇒ The possible values of 7−xPx−3 are when x = 3,4 or 5 . ⇒ The range of values of 7−xPx−3 is { 4P0, 3P1, 2P2 } or {1,3,2}. 21. (B) 18 correct results can be predicted in 21C18 ways and 3 wrong results in 2 3 ways. Thus, required number of ways is 21C182 3 . 22. (C) A, I, I, I, O can occur at odd places in 5! 3! ways, and the remaining letters N, D, L, L can be arranged at the remaining places in 4! 2! ways. 23. (A) Number of ways = 5C1 + 5C2 + 5C3 = 25 24. (D) As 0 + 1 + 2 + 3 + 4 + 5 = 15, for a five digit number either do not use 0 or 3 . When 0 is not used, the number of such numbers is 5 ! And when 3 is not used, the number of such number is 5! − 4 !. ∴ Required number of ways = 5! + (5! − 4!) = 216 25. (A) We can choose 10 players from (22 - 2) players in 20C10 ways and one player from 2 players in 2C1 ways. ∴ Required number of ways = 20! 10!10! × 2 = 369512 26. (B) Total number of books = a + 2b + 3c + d Since there are b copies of each of two books, c copies of each of three books and single copies of d books. Therefore the total number of arrangements is (a+2b+3c+d)! a!(b!) 2(c!) 3 . 27. (A) Let n = 2m + 1. Then, the common difference of the A.P. can be 1,2,3, ... , m. The number of AP's with 1,2,3, ..., m common differences are (2m − 1), (2m − 3), ... .1 respectively So, total number of AP's = (2m − 1) + ⋯ + 1 = m2 = ( n−1 2 ) 2 . Hence, total number of ways = ( n−1 2 ) 2 28. (B) Consider the host and the two particular persons as a single unit. So, there are effectively 19 persons. These 19 persons can be arranged in ⌊8 ways and there are ⌊2 ways of arranging the two persons on either side of the host. 29. (C) Digit at the extreme left can be chosen in 9 ways. ( 0 is not a possibility). Now the next digit can be chosen in 9 ways as consecutive digits are not same and so on. ⇒ Required number of numbers = 9 × 9 × ... × 9 = 9 n n times 30. (A) Required no. of ways = 22C19 = 1540 31. (B) ∣2n+1 n+2 × ∣n−1 2n−1 = 3 5 ⇒ (2n+1)(2n) (n+2)(n+1)(n) = 3 5 ⇒ n = 4 32. (B) The number of points of intersection = ( nC2 )(2) + ( mC2 )(1) + ( nC1 )( mC1 )(2) = n(n − 1) + 1 2 m(m − 1) + 2mn 33. (B) Number of ways is 12C6 for M, 6C4 for P and 2C2 for C .
Thus, the required number of ways = ( 12C6 )( 6C4 )( 2C2 ) 34. (A) The total number of points is 15 . From these 15 points we can obtain 15C3 triangles. However, if all the 3 points are chosen on the same straight line, we do not get a triangle. Therefore, the required number of triangles = 15C3 − 3( 5C3 ) = 425 35. (C) The required number of triangles is 15C3 − 4C3 = 455 − 4 = 451 36. (D) MEDITERRANEAN R → 2, M → 1,D → 1, I → 1, T → 1, A → 2, N → 2, E → 3 R−E For letter words beginning with R (first letter) and ending with E (fourth letter) can be formed by two ways. (i) All two letters between R and E are distinct. (ii) All two letters between R and E are identical. (i) Excluding one R and on E, total distinct letters are 8 (R, M, D, I, T, A, N, E). (ii) Excluding one R and E , total identical pairs are 3 (AA, NN, EE). ∴ Total such words = 56 + 3 = 59. 37. (A) nCr = n! r! (n − r)! n Pr = n! (n − r)! n+2C6 n−2P2 = 11 ⇒ (n + 2)! 6! (n + 2 − 6)! / (n − 2)! (n − 2 − 2)! = 11 ⇒ (n + 2)! (n − 2)! = 11 × 6! ⇒ (n + 2)(n + 1)n(n − 1) = 11 × 10 × 9 × 8 Equation on the both sides, n = 9 Clearly, n = 9 satisfies (A) ∵ n 2 + 3n − 108 = 0 ⇒ (n + 12)(n − 9) = 0 38. (B) Required numbers of ways = 15C1 × 15C1 + 14C1 × 14C1 + 13C1 × 13C1 + ⋯ . + 1C1 × 1C1 = 152 + 142 + 132 + ⋯ . +1 2 = 15(15 + 1)(2 × 15 + 1) 6 = 15 × 16 × 31 6 = 1240 39. (D) Here A = {x1, x2, ... . , x7 }; B = {y1, y2, y3 } f: A → B There are three elements in A such that f(x) = y2. This can be done by 7C3 ways. Out of other 4 elements of A, each one can be mapped with B except y2 by 2 ways. So, 4 elements of A can be mapped with 2 elements of B by 2 4 ways. This contains 2 mapping such that all 4 element of A can be mapped with one element of B such that f(x) = y1 or f(x) = y3, i.e. 2 ways. So, for onto functions, it must be 2 4 − 2 = 14. ∴ Total number of onto functions f: A → B is 14 × 7C3. 40. (B) Sum of the digits in the unit's place of all the 4 digit numbers = (4 − 1)! (sum of all four digits) = 3! (3 + 4 + 5 + 6) = 108 41. (D) A number is divisible by 9 if its sum of digits is divisible by 9 . 1 + 2 + 3 + ⋯ . +8 = 8 × 9 2 = 36 is div. by 9 Possible numbers with no repetitions are (i) 8 digits number excluding 09 = 8 ! (ii) 8 digits number excluding 18 = 7 × 7 ! (iii) 8 digits number excluding 27 = 7 × 7 ! (iv) 8 digits number excluding 36 = 7 × 7 ! (v) 8 digits number excluding 45 = 7 × 7! ∴ Total number = 8! + 4 × 7 × 7! = 36(7!) 42. (B) Number of games that n men played between themselves = 2 ⋅ nC2 Number of games that n men played with 2 women = 2 ⋅ 2n ∴ 2 ⋅ nC2 − 2 ⋅ 2n = 66 ⇒ n(n − 1) − 4n = 66 ⇒ (n − 11)(n + 6) = 0 ∴ n = 11 43. (C) x1 + x2 + x3 + ⋯ + x8 = 30, xi ≥ 2 Let xi = ai + 2 then a1 + a2 + a3 + ⋯ . +a8 + 16 = 30 a1 + a2 + a3 + ⋯ + a8 = 14, ai ≥ 0 Total number of ways of dividing n identical items among r persons, each of whom, can receive 0,1,2,3 .... or more items (≤ n) = n+r−1Cr−1 ∴ Required number of ways = 14+8−1C8−1 = 21C7 44. (365) (0,2,4,6,8) Numbers not containing 5 8 + × 9 + × 5 + 8 × 9 × 5 = 360 (First digit should not be 0 or 5,2 nd digit should not be 5, 3 rd digit should be even)) 45. (8) ab should both be at odd places or even places. 2 × (2!)(2!) = 8
46. (162) 1 st , 2 nd identical. 3 rd should not be identical 9 × 9 = 81 2 nd , 3 nd , identical 1 st not identical \non-zero. 9 × 9 = 81 Total = 81 + 81 = 162 47. (41) Let two particular persons who either serve together or not at all are A and B. Also, let two particular persons who refuse to serve with each other are C and D. Case 1: AB in = 7C3 − 5C1 = 35 − 5 = 30 AB in −AB in and CD also in Case 2: AB out = 7C5 − 5C3 = 11 AB out −AB out and CD in 48. (126) There is only one arrangement of 5 different numbers in ascending or descending order. For a number to be a 5 digit zero should not be there while arranging in ascending order. m = 10C5, n = 9C5 49. (80) Each should belong to different pair and each shoe can have 2 options left or right shoe. 5C4 × 2 4 = 5 × 16 = 80 50. (240) 1̅, 2̅, 3̅, 3̅ 41 21 4 or 5 ⏟−̅ Sum =9 51. (52) (1) Exactly one 2 's \one 3's ( 3 rd digit has 8 options ) − (3 nd digit is a zero and no. begins with 0) 8C1 × 3! − 2! = 8 × 6 − 2 = 46 (2) Exactly two 2's \one 3's or one 2's or two 3's. 2 × ( 3! 2! ) = 6 46 + 6 = 52 52. (126) 5 match series I I I I I = 1 = 4C4 (Last match won by India and 4 out of 1 st 5 won by India) [Here I at any position denotes that particular match is won by India] 7 match series 6C4 8 match series ... 7C4 Total 4C4 + 5C4 + 6C4 + 7C4 + 8C4 = 126 53. (8) 3 × 2 n−1 = 384 2 n−1 = 128 = 2 7 n − 1 = 7 ⇒ n = 8 54. (10) Let n(A), n(B), n(C) denote number of people possessing TV, VCR and tape recorder respectively \(A, B, C) be respectively set of teachers possessing them. n(A) = 22, n(B) = 15, n(C) = 14 n(A ∩ B ∩ C) = 1 Exactly two = n(A ∩ B) + n(B ∩ C) + n(C ∩ A) − 3n(A ∩ B ∩ C) 9 = n(A ∩ B) + n(B ∩ C) + n(C ∩ A) − 3 × 1 ∴ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 12 Number of teachers possessing at least one = n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − (n(A ∩ B) + n(B ∩ C) + n(C ∩ A)) + n(A ∩ B ∩ C) = 22 + 15 + 14 − 12 + 1 = 40 Number of teachers possessing none = 50 − 40 = 10 55. (144) Total ways in AB are adjacent - Total ways in which AB are adjacent and CD are adjacent. AB, C, D, E, F 5! × 2! = 240 − 4! × 2! × 2! = 96 56. (30) last 2 digits should be 04,12,20,24,32,40 Ending with 04,20,40 = 3 × (3!) = 18 Ending with 12,24,32 = 3 × (31 − 2 ! ) Starting with O . 3 × (6 − 2) = 3 × 4 = 12 57. (50) Let A, B, C be the sets of students who read business India, business World and business Today.