1. Mathematics

2. Permutation & Combination Session

4. Session Objective 1.Factorial 2.Fundamental principles of counting 3.Permutations as arrangement 4. n P r formula 5.Permutations under conditions 6.Permutation of n objects taken r at a time

5. Permutation – Its arrangement Two element – ab Arrangements  (a, b), (b, a) = 2 Father Son Father is riding Father Son Father is teaching

6. Permutation – Its arrangement Three elements a, b, c Arrangements : a b c a c b b a c b c a c a b c b a  First Second Third Place Place Place

7. Permutation – Its arrangement I st 2 nd 3 rd 3 ways 2 ways 1 ways Total ways = 3 + 2 + 1 ? or 3 x 2 x 1 ?

8. Permutation – Its arrangement I st 2 nd 2 ways 1 ways x Total ways = 2 + 1 = 3 or 2 x 1 = 2

9. Permutation – Its arrangement I st 2 nd 3 rd 3 ways 2 ways 1 ways x Total ways = 3 + 2 + 1 ? or 3 x 2 x 1 ?

10. Number of Modes A  B? Cycle Scooter Car B A 1 Km Number of modes to reach B = Add/Multiply 1 1 Bus Scooter CarWalking Number of ways A  B? I II B A No. of ways = 1 + 1 = 2 Independent Process + + + = 4

11. Mode & Way Number of style A  B? I II B A Scooter Cycle Car Ways – 2 Modes – 4 To reach B, one dependent on both ways and mode. Number of style = 4 x 2 = 8 Independent Process  + Dependent Process  X

12. Mode & Way There are two ways and 4 modes for A  B. How many way one can reach B from A? One can reach Lucknow from New Delhi only through Kanpur (No direct root) I II III IV A B Kanpur Lucknow New Delhi Process  Dependent I A I B II A II B III A III B IV A IV B No. of ways = 4 x 2 = 8

13. Mode & Way I II III IV A B Kanpur Lucknow New Delhi IV V

14. Questions

15. Illustrative Problem From the digits 1, 2, 3, 4, 5 how many two digit even and odd numbers can be formed. Repetition of digits is allowed. Solution: Total nos = 5 Even number 5 ways2 ways (2/4) Even numbers=5 x 2=10

16. Solution contd.. Odd number 5 ways 3 ways (1/3/5) Odd numbers = 5 x 3 = 15 Total numbers 5 ways Total numbers = 5 x 5 = 25 Even Numbers=10

17. Illustrative Problem From the digits 1, 2, 3, 4, 5 how many two digit numbers can be formed. When repetition is not allowed. Solution: 5 ways 4 ways Total = 5 x 4 = 20

18. Illustrative Problem There are three questions. Every question can be answered in two ways, (True or False). In how many way one can answer these three questions? Solution: 2 ways (T/F) 2 ways Question I st 2 nd 3 rd 2 ways No. of ways = 2 x 2 x 2 = 8 T T F T F T F F T F T F T F

19. Illustrative Problem In a class there are 10 boys and 8 girls. For a quiz competition, a teacher how many way can select (i) One student (ii) One boy and one girl student (i)Independent of whether boy/girl = 10 + 8 = 18 ways Solution: (ii)Dependent process = 10 x 8 = 80 ways 8ways 10 ways GirlBoy

20. 8 10 Solution : (iii) Girl Boy1 Boy2 (iv)student=10+8=18 In a class there are 10 boys and 8 girls. For a quiz competition, a teacher how many way can select (iii) two boys and one girl (iv) two students 9 10 x 9 x 8 = 720 Illustrative Problem =18 x 17 18 17 student1student2

21. In a class there are 10 boys and 8 girls. For a quiz competition, a teacher how many way can select (v) at least one girl while selecting 3 students Solution: case1: 1 girl 8 10 9 10 x 9 x 8 = 720 boys-10 girl-8 G BB case2: 2 girl 8 7 10 8 x 7 x 10 = 560 G GB case3: 3 girl 8 x7 x 6 = 186 Ans: 720+560+186=1666 Illustrative Problem

22. Principle of Counting Multiplication Principle : If a job can be done in ‘m’ different ways, following which another can be done in ‘n’ different ways and so on. Then total of ways doing the jobs = m x n x …… ways. Addition Principle : If a job can be done in ‘m’ different ways or ‘n’ different ways then number of ways of doing the job is (m + n). Multiplication – Dependent Process Addition – Independent Process

23. Questions

24. Illustrative Problem Eight children are to be seated on a bench. How many arrangements are possible if the youngest and eldest child sits at left and right corner respectively. Solution: We have 6 children to be seated 6 5 Youngest Eldest 4 3 2 1 No. ways = 6 x 5 x 4 x 3 x 2 x 1 = 720 x 2 = 1440 ways

25. Illustrative Problem A class consists of 6 girls and 8 boys. In how many ways can a president, vice president, treasurer and secretary be chosen so that the treasurer must be a girl and secretary must be a boy. (Given that a student can’t hold more than one position) Solution : Girls – 6 Boys - 8 Treasurer (Girl) 6 ways Girls – 5 Boys - 8 Secretary (Boy) 8 ways

26. Solution contd.. Girls – 5 Boys – 7; Total = 12 12 ways PresidentVice President 11 ways Total = 6 x 8 x 12 x 11 Treasurer(Girl)-6 ways Secretary(Boy)-8 ways

27. Factorial Defined only for non-negative integers Denoted as n ! or. n N ! = n. (n - 1). (n – 2) …… 3. 2. 1. Special case : 0 ! = 1 Example : 3 ! = 3. 2. 1 = 6 5 ! = 5. 4. 4. 3. 2. 1 = 120 (4.5) ! - Not defined (-2) ! - Not defined

28. Questions

29. Illustrative Problem Solution :

30. Illustrative Problem Find x if 8! x! = (x + 2)! 6! Solution : Short cut

31. Permutation Arrangement of a number of object(s) taken some or all at a fine. Example : Arrangement of 3 elements out of 5 distinct elements =

32. nPrnPr Arrangement of r elements out of n given distinct elements. n (n - 1) 1 st 2 nd r th (n – r + 1) ? …….

33. nPrnPr For r = n Arrangements of n distinct element n P n = n! taken all at a time.

34. Questions

35. Illustrative Problem In how many way 4 people (A, B, C, D) can be seated (a) in a row (b) such that Mr. A and Mr. B always sit together Solution : (a) 4 P 4 = 4! (b)(A, B), treat as one C, D

36. Solution contd.. Arrangement among 3 = 3 P 3 (A, B, C, D) (B, A, C, D) 2P22P2 (A, B, D, C) (B, A, D, C) 2P22P2 3P33P3 (A, B), C, D (A, B), D, C C, (A, B), D C, D, (A, B) D, (A, B), C D, C, (A, B) 3 P 3 x 2 P 2 (b)(A, B), C, D = 3!2! = 12

37. Solution contd.. In how many way 4 people (A, B, C, D) can be seated (c) A,B never sit together = Total no. of arrangement – No. (A, B) together = 4 P 4 – 3 P 3. 2 P 2 = 4! - 3! 2! = 24 – 12 = 12

38. Illustrative Problem Seven songs (Duration – 4, 4, 5, 6, 7, 7, 7, 7 mins.) are to be rendered in a programme (a) How many way it can be done (b) such that it occurs in ascending order (duration wise) Solution : (a) 7 P 7 = 7! (b) Order – 4, 4, 5, 6, 7, 7, 7 mins 2P22P2 3P33P3  No. of way = 2 P 2 x 3 P 3 = 2! 3! = 24

39. Illustrative Problem How many four digits number can be formed by the digits. 3, 4, 5, 6, 7, 8 such that (a) 3 must come (b) 3 never comes (c) 3 will be first digit (d) 3 must be there but not first digit

40. Solution (a) 3, 4, 5, 6, 7, 8 No. of 4 digit numbers with 3 = 4 x 5 P 3 ‘3’ can take any of four position. In each cases. 5 digits to be arranged in 3 position.

41. Solution contd.. (b) Digits available – 5 (4, 5, 6, 7, 8 No. of 4 digit numbers without 3 = 5 P 4 (c) 3 _ _ _ No. of digits available = 5 No. of position available = 3 No. of 4 digit number start with ‘3’ = 5 P 3

42. Solution contd.. (d) 4 digit nos. contain ‘3’ but not at first = 4 digit number with ‘3’ – 4 digit number with ‘3’ at first = solution (a) – solution (c) = 4. 5 P 3 – 5 P 3 = 3. 5 P 3

43. Illustrative Problem In how many way a group photograph of 7 people out of 10 people can be taken. Such that (a) three particular person always be there (b) three particular person never be there (c) three particular always be together Solution: (a)3 particular 7 places  7 P 3 With each arrangement X_X _ X _ _ Arrangements

44. Solution contd.. Arrangements Person available – 7 Places available – 3 7P47P4 Total no. of arrangements = 7 P 3 x 7 P 4 Person available = 7 Places available = 7 7P77P7 arrangements

45. Solution contd.. (c) X X X _ _ _ _ Treat as one No. of person = 10 – 7 + 1 = 8 Place available = 5 of which one (3 in 1) always be there.  No. of arrangement = 5. 7 P 4 3 Particular can be arranged = 3 P 3 way  Total arrangement = 5. 7 P 4. 3 P 3

46. Illustrative Problem How many way, 3 chemistry, 2 physics, 4 mathematics book can be arranged such that all books of same subjects are kept together. Solution:

47. Solution contd.. Inter subject arrangement Phy Chem Maths  Arrangement = 3 P 3 Total no. of arrangements = 3 P 3 (3p3 x 2 p 2 x 4 p 4 )

48. Class Test

49. Class Exercise - 1 If are in the ratio 2 : 1, find the value of n.

50. Solution Given that Answer is n = 5 (rejecting n = 0).

51. Class Exercise - 2 How many numbers are there between 100 and 1000 such that each digit is either 3 or 7?

52. Solution By fundamental principle of counting, the required number = 2 × 2 × 2 = 8 (Each place has two choices.)

53. Class Exercise - 3 How many three-digit numbers can be used using 0, 1, 2, 3 and 4, if (i)repetition is not allowed, and (ii)repetition is allowed?

54. Solution (i) Hundred’s place can be filled in 4 ways. Ten’s place can be filled in 4 ways. Unit’s place can be filled in 3 ways. Required number = 4 × 4 × 3 = 48 (ii) Similarly, the required number = 4 × 5 × 5 = 100

55. Class Exercise - 4 How many four-digit numbers have at least one digit repeated?

56. Solution Number of four-digit numbers = 9 × 10 × 10 × 10 = 9000 Number of four-digit numbers with no repetition = 9 × 9 × 8 × 7 = 4536 Number of four-digit numbers with at least one digit repeated = 9000 – 4536 = 4464

57. Class Exercise - 5 There are 5 periods in a school and 6 subjects. In how many ways can the time table be drawn for a day so that no subject is repeated?

58. Solution Six subjects can be allocated to five periods in ways, without a subject being repeated.

59. Class Exercise - 6 Number of ways in which 7 different sweets can be distributed amongst 5 children so that each may receive at most 7 sweets is (a)7 5 (b) 5 7 (c) 7 p 5 (d) 35

60. Solution Each sweet can be given to any of the 5 children. Thus, the required number is 5 × 5 × 5 × 5 × 5 × 5 × 5 = 5 7 Hence answer is (b)

61. Class Exercise - 7 In how many ways can 5 students be seated such that Ram always occupies a corner seat and Seeta and Geeta are always together?

62. Solution Seeta and Geeta can be arranged in 2 ways. Remaining students can be arranged in 2 ways. Total ways = 2 × 3 × 2 × 2 = 24 Ram can be seated in 2 ways. Seeta and Geeta can be together in 3 ways. (If Ram occupies seat 1, Seeta-Geeta can be in 2-3, 3-4 or 4-5.)

63. Class Exercise - 8 How many words can be made from the word ‘helicopter’ so that the vowels come together?

64. Solution Treating the vowels as one unit, we have 7 units. These can be arranged in 7! ways. The vowels can be arranged in 4! ways. Total ways = 7! × 4! = 120960

65. Class Exercise - 9 There are 5 questions. Each question has two options (one answer is correct). In how many ways can a student fill up the answer sheet, when he is asked to attempt all the questions?

66. Solution Every question can be answered in two ways. = 2 ×2 × 2 × 2 × 2 = 2 5 = 32 ways. Five questions can be answered in

67. Class Exercise - 10 In how many ways can 6 students (3 boys and 3 girls) be seated so that (i)the boys and girls sit alternatively, (ii)no 2 girls are adjacent?

68. Solution (i)Case 1: From left-side, when first student is a boy, then the boys can occupy Ist, 3rd and 5th places. And the girls can occupy 2nd, 4th and 6th places. So the boys can be seated in 3 p 3 ways and the girls can be seated in 5 p 3 ways. Number of arrangement = 3!. 3!

69. Solution contd.. Case 2: When the first student is a girl (from left), then also the number of permutation = 3! × 3! Therefore, total number of permutation = 2 × (3!) 2 = 72 (ii) Let first boys are seated. They can sit in three places in 3 p 3 = 3! ways. Since no girls should be adjacent, the number of seats left for girls are four. __ B __ B __ B __ Number of permutation for girls = 4 p 3 = 4! Therefore, total number of permutation = 3! 4! = 144

70. Thank you