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Mathematics

Mathematics. Session. Permutation & Combination. Session Objective. 1. Factorial 2. Fundamental principles of counting 3. Permutations as arrangement 4. n P r formula Permutations under conditions Permutation of n objects taken r at a time. Father. Father. Son. Son. Father is riding.

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Mathematics

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  1. Mathematics

  2. Session Permutation & Combination

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

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

  5. Permutation – Its arrangement Three elements a, b, c First Second ThirdPlace Place Place Arrangements : a b c a c b b a c = 6 b c a c a b c b a

  6. 3 ways 2 ways 1 ways Total ways = 3 + 2 + 1 ? or 3 x 2 x 1 ? Permutation – Its arrangement Ist 2nd 3rd

  7. 2 ways 1 ways Total ways = 2 + 1 = 3 or 2 x 1 = 2 Permutation – Its arrangement Ist 2nd x

  8. 3 ways 2 ways 1 ways Total ways = 3 + 2 + 1 ? or 3 x 2 x 1 ? Permutation – Its arrangement Ist 2nd 3rd x

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

  10. Scooter Cycle Car I B A II Mode & Way Number of style A  B? 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

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

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

  13. Questions

  14. 5 ways 2 ways (2/4) Illustrative Problem From the digits 1, 2, 3, 4, 5 how many twodigit even and odd numbers can be formed.Repetition of digits is allowed. Solution: Total nos = 5 Even number Even numbers=5 x 2=10

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

  16. 5 ways 4 ways Illustrative Problem From the digits 1, 2, 3, 4, 5 how many twodigit numbers can be formed. Whenrepetition is not allowed. Solution: Total = 5 x 4 = 20

  17. T T F T T F F 2 ways(T/F) 2 ways 2 ways T T F F T F F Illustrative Problem There are three questions. Every question canbe answered in two ways, (True or False). Inhow many way one can answer these threequestions? Solution: Question Ist 2nd 3rd No. of ways = 2 x 2 x 2 = 8

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

  19. 8 18 10 17 9 Illustrative Problem In a class there are 10 boys and 8 girls.For a quiz competition, a teacher howmany way can select(iii) two boys and one girl(iv) two students Solution : (iii) Girl Boy1 Boy2 10 x 9 x 8 = 720 (iv)student=10+8=18 student1 student2 =18 x 17

  20. 8 8 10 7 10 9 Illustrative Problem In a class there are 10 boys and 8 girls.For a quiz competition, a teacher howmany way can select(v) at least one girl while selecting 3 students case1: 1 girl Solution: G B B boys-10 girl-8 10 x 9 x 8 = 720 G G B case2: 2 girl 8 x 7 x 10 = 560 case3: 3 girl 8 x7 x 6 = 186 Ans: 720+560+186=1666

  21. Principle of Counting Multiplication Principle : If a job can bedone in ‘m’ different ways, following whichanother can be done in ‘n’ different ways andso on. Then total of ways doing the jobs= m x n x …… ways. Addition Principle : If a job can bedone in ‘m’ different ways or ‘n’ different waysthen number of ways of doing the job is (m + n). Multiplication – Dependent Process Addition – Independent Process

  22. Questions

  23. 5 4 3 2 1 6 Illustrative Problem Eight children are to be seated on a bench.How many arrangements are possible if theyoungest and eldest child sits at left and rightcorner respectively. Solution: We have 6 children to be seated Youngest Eldest No. ways = 6 x 5 x 4 x 3 x 2 x 1 = 720 x 2 = 1440 ways

  24. Treasurer (Girl) Secretary (Boy) 6 ways 8 ways Illustrative Problem A class consists of 6 girls and 8 boys. In howmany ways can a president, vice president,treasurer and secretary be chosen so thatthe treasurer must be a girl and secretarymust be a boy. (Given that a student can’thold more than one position) Solution : Girls – 6 Boys - 8 Girls – 5 Boys - 8

  25. 12 ways 11 ways Solution contd.. Treasurer(Girl)-6 ways Secretary(Boy)-8 ways Girls – 5 Boys – 7;Total = 12 President Vice President Total = 6 x 8 x 12 x 11

  26. Defined only for non-negative integers • Denoted as n ! or . n Factorial 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

  27. Questions

  28. Illustrative Problem Solution :

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

  30. Permutation Arrangement of a number of object(s) takensome or all at a fine. Example : Arrangement of 3 elements out of5 distinct elements =

  31. rth ? n (n - 1) nPr Arrangement of r elements out of n givendistinct elements. 1st 2nd ……. (n – r + 1)

  32. nPr For r = n Arrangements of n distinct element nPn = n!taken all at a time.

  33. Questions

  34. (A, B), treat as one 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) 4P4 = 4! (b) C, D

  35. (A, B), C, D (A, B), D, C C, (A, B), D (A, B, C, D) (A, B, D, C) C, D, (A, B) 2P2 2P2 (B, A, D, C) (B, A, C, D) D, (A, B), C D, C, (A, B) Solution contd.. (b) (A, B), C, D Arrangement among 3 = 3P3 3P3 3P3 x2P2 = 3!2! = 12

  36. 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 = 4P4 – 3P3.2P2 = 4! - 3! 2! = 24 – 12 = 12

  37. (b) Order – 4, 4, 5, 6, 7, 7, 7 mins 2P2 3P3 Illustrative Problem Seven songs (Duration – 4, 4, 5, 6, 7,7, 7, 7 mins.) are to be rendered in aprogramme(a) How many way it can be done(b) such that it occurs in ascending order (duration wise) Solution : (a) 7P7 = 7!  No. of way = 2P2 x 3P3 = 2! 3! = 24

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

  39. Solution (a) 3, 4, 5, 6, 7, 8 No. of 4 digit numbers with 3 = 4 x 5P3 ‘3’ can take any of four position. In each cases. 5 digits to be arranged in 3 position.

  40. Solution contd.. (b) Digits available – 5 (4, 5, 6, 7, 8 No. of 4 digit numbers without 3 = 5P4 (c) 3 _ _ _ No. of digits available = 5 No. of position available = 3 No. of 4 digit number start with ‘3’ = 5P3

  41. 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) = 4.5P3 – 5P3 = 3.5P3

  42. Illustrative Problem In how many way a group photograph of7 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: Arrangements • 3 particular 7 places  7P3 • With each arrangement • X _ X _ X _ _

  43. Person available – 7 Places available – 3 7P4 Person available = 7 Places available = 7 7P7 arrangements Solution contd.. Arrangements Total no. of arrangements = 7P3 x 7P4

  44. (c) XXX _ _ _ _ Treat as one Solution contd.. No. of person = 10 – 7 + 1 = 8Place available = 5of which one (3 in 1) always be there.  No. of arrangement = 5.7P43 Particular can be arranged = 3P3 way Total arrangement = 5.7P4.3P3

  45. Illustrative Problem How many way, 3 chemistry, 2 physics,4 mathematics book can be arranged suchthat all books of same subjects are kepttogether. Solution:

  46. Phy Chem  Arrangement = 3P3 Maths Solution contd.. Inter subject arrangement Total no. of arrangements = 3P3(3p3 x 2p2 x 4p4)

  47. Class Test

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

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

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