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Mathematics

Mathematics. Session. Permutation & Combination - 2. Session Objectives. Session Objective. 1. Combination. 2. Circular Permutation. Selection. Rejection. Selection. Rejection. Combination. Combination  Selection. Selection from a, b, c. a , b , c ,. b , a , c ,. Select one.

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Mathematics

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  1. Mathematics

  2. Session Permutation & Combination - 2

  3. Session Objectives

  4. Session Objective 1. Combination 2. Circular Permutation

  5. Selection Rejection Selection Rejection Combination Combination  Selection Selection from a, b, c a , b , c , b , a , c , Select one c , a , b , a , b, c , b , c , a , No. of ways = 3 Select two c , a , b ,

  6. Selection Rejection a, b, c Combination Select three No. of ways = 1 Number of selection of some from a group. = Number of rejection of remaining.

  7. Selection Rejection S1 S2 S3 S4 S1 S3 S2 S4 S1 S4 S2 S3 S2 S3 S1 S4 S2 S4 S1 S3 S3 S4 S1 S2 Combination Number of ways of selecting a group of two student out of four for a trip to Goa. S 1, S 2, S 3, S 4 Select two 6 ways.

  8. Combination Number of ways of selecting one group Of two for Goa other for Agra

  9. Combination Selection and Arrangement of 3 alphabets from A, B, C, D.

  10. Number of distinct elements = n (1,2,3, .. n). Ways of rejecting r = nCr Ways of rejecting rest (n – r) elements = nCn-r nCr= nCn-r Particular selection 1,3, … r elements Arrangement r! Total no. of arrangement = nCr.r! = nPr Combination

  11. Questions

  12. There are 5 man and 6 woman. How many way one can select (a) A committee of 5 person. (b) A committee of 5 which consist exactly 3 man. (c) A committee of 5 persons which consist at least 3 man. Illustrative Problem Solution : Man – 5 Woman – 6 Total - 11

  13. Solution Cont. (c) At least – 3 man Man – 5 Woman - 6 Composition of Committee Case Man Woman 3 2 5C3 x 6C2 = 150 4 1 5C4 x 6C1 = 30 5 0 5C5 x 6C0 = 1 No. of Ways = 150 + 30+ 1 = 181.

  14. In how many ways, a committee of 4 person Can be selected out of 6 person such that (a) Mr. C is always there (b) If A is there B must be there. (c) A and B never be together. No. of persons - 6 Committee - 4 (b) Case – 1 : ‘ A is there’ – ‘AB’ in Committee. Available persons – 4 Persons to select - 2 Ways = 4C2 Illustrative Problem Solution : • Available persons – 5 • Persons to select - 3 Ways = 5C3

  15. Available persons – 5 Persons to select - 4 Ways = 5C4 Solution Cont. Case – 2 – ‘A is not there’ – B may /may not be there No. of person - 6 Person to Select - 4 (c) ‘AB’ never together = total no. of committee - ‘AB’ always together. Total no. of committee = 6C4 . ‘AB together in committee’ = 4C2  No. of Ways = 6C4 – 4C2 = 9

  16. How many straight lines can be drawn through 15 given points. when (a) No. three are collinear (b) Only five Points are collinear Illustrative Problem Solution : Through two given point and unique straight line

  17. Form 4 digit nos. using these three 5 digit Select 3 distinct Number of digits formed. Select one which will repeat 3 digit Illustrative Problem Find the number of 4 digit numbers that can be formed by 3 distinct digits among 1,2,3,4,5 No. of digits = 5 Solution :- 5C3

  18. Illustrative Problem In how many ways 9 students can be seated both sides of a table having 5 seat on each side (non-distinguishable)

  19. abcd dabc cdab bcda a 4 linear arrangement b d 1 circular arrangement c Circular Permutation A,B,C,D – to be seated in a circular table Total line arrangement = 4! 4 linear Arrangement  1 circular arrangement 4 ! linear Arrangement  No. of circular arrangement of n object = (n-1) !

  20. Question

  21. In how many way 4 girls and 5 boys can be seated around a circular table such that (i) No. two girls sit together (ii) All girls sit together (iii) Only two girls sit together Illustrative Problem

  22. B1 B2 B5 B3 B4 Solution (i) Boys – 5 Girls – 4

  23. G’s 4! B1 B2 B5 B3 B4 Solution Cont. (ii) Boys – 5 Girls – 4

  24. 2! B1 G3G4 B2 B3 B5 G2 G1 B4 Solution Cont. (iii) Boys – 5 Girls – 4

  25. B B A C C D D No. of arrangement = Invertible Circular Arrangement Ex :- Garland, Neck less. Clockwise and anticlockwise arrangement considered as same For n objects.

  26. Question

  27. Step 1 :- choose 5 Illustrative Problem I have ten different color stones. In How many way I can make a ring of five stones Solution : Stone - 10 Step 2 :- arrange circularly (Invertible)

  28. 100th place 10th place Unitplace 3 1 2 2 1 3 3 2 1 1 2 3 2 3 1 1 3 2 12 12 12 Sum of Digits Find the sum of all three digit numbers formed by 1, 2 and 3 Sum of digits same (12) Each digits is repeated same no. of times =2=(3-1) ! All digits comes equal no. of times Sum of digits in each column = (1+2+3) x 2! = 12

  29. Unitplace 10th place 100th place b c a 12 12 12 Sum of Digits = 100a +10b +c = 100x12 +10x12 + 12 = 12 (102 + 10 + 1) = 12 x 111 = 1332 Sum of all numbers = (sum of all digits) x (No. of repetition in a particular column) x (No. of 1’s as number of digits present in the number

  30. Class Test

  31. Class Exercise - 1 In how many ways can 5 boys and5 girls be seated in a row so that no2 girls are together and at least2 boys are together?

  32. Solution First the boys can be seated in 5p5 = 5! ways. Each arrangement will create six gaps: __ B __ B __ B __ B __ B If the girls are seated in the gaps, no 2 girls will be together. Girls can be seated in the gaps in 6p5= 6! ways. But if the girls are in the first five or the last five gaps, no 2 boys will be together. So the girls can be seated in 6! – (5! + 5!)= 6! – 2 × 5!= 4 × 5!

  33. Solution contd.. Thus, total arrangements possible are 4 x (5!)2 = 4 × 120 × 120 = 57600 Note: Under the given condition, more than 2 boys cannot sit together.

  34. Class Exercise - 2 Find the sum of all numbers formedusing the digits 0, 2, 4, 7.

  35. Required sum is Solution = 16 × 13 × 4333 = 208 × 4333 = 901264

  36. Class Exercise - 3 If all the letters of the word‘SAHARA’are arranged as in the dictionary, whatis the 100th word?

  37. Number of words starting with A: Number of words starting with H: Number of words starting with R: Solution Arranging the letters alphabetically, we haveA, A, A, H, R, S. Thus, the last word starting with R will be the100th word. This is clearly RSHAAA.

  38. Class Exercise - 4 How many numbers can be formedusing the digits 3, 4, 5, 6, 5, 4, 3such that the even digits occupythe even places?

  39. These can be arranged in the even placesin ways = 3 ways. The remaining digits: 3, 5, 5, 3 can be arranged in the remaining places in ways. Solution Even digits are 4, 6, 4. Thus, the total number of ways = 3 × 6 = 18 ways.

  40. Class Exercise - 5 Ten couples are to be seatedarounda table. In how many ways can theybe seated so that no two neighboursare of the same gender?

  41. Solution Let all the members of one gender be seated around the table. This can be done in (10 – 1)! ways. Once one gender is seated, arrangement of other gender is no longer a problem of circular permutation (since the seats can be identified). Thus, the second gender can be seated in 10! ways. Thus, total ways = 9! × 10!

  42. Class Exercise - 6 In how many ways can 15 delegatesbe seated around a pentagonal tablehaving 3 chairs at each edge?

  43. Solution If we consider the problem as one of circular permutations, the answer is (15 – 1)! = 14! But we are considering the above two arrangements as same while they are clearly different. All that has been done is that all delegates have shifted one position. One move shift will also give a new arrangement.

  44. But after three shifts, the arrangement will be which is identical to the original arrangement. or, where 5 is the number of sides of regular polygon. Solution contd.. Thus, we are counting three different arrangements as one. Thus, number of actual arrangements possible = 3 × 14!

  45. Class Exercise - 7 Prove that the product of r consecutiveintegers is divisible by r!

  46. Solution Let the r consecutive integers be (n + 1), (n + 2), (n + 3), ..., (n + r) Product = (n + 1)(n + 2)(n + 3) ... (n + r) But n+rPris an integer. Thus, the product of r consecutive integers is divisible by r! (Proved)

  47. If find the values of n and r. Class Exercise - 8

  48. Solution

  49. Class Exercise - 9 A person wishes to make up as manyparties as he can out of his 18 friendssuch that each party consists of thesame numbers of persons. How manyfriends should he invite?

  50. Solution Let the person invite r friends. This canbe done in 18Crways. To maximise thenumber of parties, we have to take thelargest value of 18Cr. When n is even,nCr will be maximum when r= n/2. Thus, he should invite 18/2 = 9 friends.

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