1. Mathematics

2. Permutation & Combination - 2 Session

3. Session Objectives

4. Session Objective 1.Combination 2.Circular Permutation

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

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

7. 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 SelectionRejection S 1 S 2 S 3 S 4 S 1 S 3 S 2 S 4 S 1 S 4 S 2 S 3 S 1 S 4 S 2 S 4 S 1 S 3 S 3 S 4 S 1 S 2 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. Selection ArrangementRejection A, B, C,ABC, ACB, BAC, BCA, CAB, CBAD A, B, D,ABD, ADB, BAD, BDA, DAB, DBAC B, C, D,BCD, BDC, CBD, CDB, DBC, DCBA A, C, D,ACD, ADC, CAD, CDA, DAC, DCAB

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

11. Questions

12. Illustrative Problem 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. Solution : Man – 5Woman – 6 Total - 11

13. Solution Cont. (c) At least – 3 man Man – 5Woman - 6 Composition of Committee Case ManWoman 32 5 C 3 x 6 C 2 = 150 41 5 C 4 x 6 C 1 = 30 50 5 C 5 x 6 C 0 = 1 No. of Ways = 150 + 30+ 1 = 181.

14. Illustrative Problem 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. Solution : No. of persons - 6Committee - 4 (a)Available persons – 5 Persons to select - 3 Available persons – 4 Persons to select - 2 Ways = 4 C 2 (b) Case – 1 : ‘ A is there’ – ‘AB’ in Committee. Ways = 5 C 3

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

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

17. Illustrative Problem Find the number of 4 digit numbers that can be formed by 3 distinct digits among 1,2,3,4,5 Solution :- No. of digits = 5 5 digitSelect 3 distinct Form 4 digit nos. using these three 3 digit Select one which will repeat Number of digits formed. 5C35C3

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

20. Question

21. Illustrative Problem 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

22. Solution (i) Boys – 5Girls – 4 B1B1 B2B2 B3B3 B4B4 B5B5

23. Solution Cont. (ii) Boys – 5Girls – 4 B1B1 B2B2 B3B3 B4B4 B5B5 4! G’s

24. Solution Cont. (iii) Boys – 5Girls – 4 B1B1 B2B2 B3B3 B4B4 B5B5 2! G2G2 G1G1 G3G4G3G4

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

26. Question

27. 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) Step 1 :- choose 5

28. Sum of Digits Find the sum of all three digit numbers formed by 1, 2 and 3 100 th place10 th placeUnit place 12 3 13 2 21 3 23 1 31 2 32 1 12 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. Sum of Digits 100 th place 10 th place Unit place a bc 12 = 100a +10b +c = 100x12 +10x12 + 12 = 12 (10 2 + 10 + 1) = 12 x 111= 1332 (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 Sum of all numbers =

30. Class Test

31. Class Exercise - 1 In how many ways can 5 boys and 5 girls be seated in a row so that no 2 girls are together and at least 2 boys are together?

32. Solution First the boys can be seated in 5 p 5 = 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 6 p 5 = 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 formed using the digits 0, 2, 4, 7.

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

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

37. Solution Arranging the letters alphabetically, we have A, A, A, H, R, S. Number of words starting with A: Number of words starting with H: Number of words starting with R: Thus, the last word starting with R will be the 100 th word. This is clearly RSHAAA.

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

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

40. Class Exercise - 5 Ten couples are to be seated around a table. In how many ways can they be seated so that no two neighbours are 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 delegates be seated around a pentagonal table having 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. Solution contd.. Thus, we are counting three different arrangements as one. Thus, number of actual arrangements possible = 3 × 14! 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.

45. Class Exercise - 7 Prove that the product of r consecutive integers 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+r P r is an integer. Thus, the product of r consecutive integers is divisible by r! (Proved)

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

48. Solution

49. Class Exercise - 9 A person wishes to make up as many parties as he can out of his 18 friends such that each party consists of the same numbers of persons. How many friends should he invite?

50. Solution Let the person invite r friends. This can be done in 18 C r ways. To maximise the number of parties, we have to take the largest value of 18 C r. When n is even, n C r will be maximum when r= n/2. Thus, he should invite 18/2 = 9 friends.

51. Class Exercise - 10 In how many ways can a cricket team of 5 batsmen, 3 all-rounders, 2 bowlers and 1 wicket keeper be selected from 19 players including 7 batsmen, 6 all-rounders, 3 bowlers and 3 wicket keepers?

52. Solution The batsmen can be selected in 7 C 5 = 21 ways. The all-rounders can be selected in 6 C 3 = 20 ways. The bowlers can be selected in 3 C 2 = 3 ways. The wicketkeeper can be selected in 3 C 1 = 3 ways. Thus, the total ways of selecting the team = 21 × 20 × 3 × 3 = 3780 ways. Note:

53. Class Exercise - 11 In how many ways can we select one or more items out of a, a, a, b, c, d, e?

54. Solution We can select ‘a’s in 0 or 1 or 2 or 3, i.e. in 4 ways. We can select ‘b’ in 2 ways, i.e. either we select it or we do not select it and so on. The required number of ways in which we can select one or more items is (3 + 1)(1 + 1)(1 + 1)(1 + 1)(1 + 1) – 1

55. Class Exercise - 12 In how many ways can we divide 10 persons (i)into groups of 5 each, (ii)(ii) into groups of 4, 4 and 2?

56. Solution We have divided by 2!, because if we interchange persons in group one, with persons in group two, the division is not different, i.e. group 1group 2group 2group 1 abdfj ceghi ceghiabdfj

57. Thank you