1. Permutation and Combination
SRI HARSHA

2. Basic Information
Factorial: The continued Product of first n natural numbers is called n! n! = 1*2*3*4*…… (n–1) * n 0! = 1 Formulae: nPr = n! / (n – r)! nCo = nCn = 1 , nC1 = n Tips: If ‘n’ is even then the greatest value of nCr is nCn/2. If ‘n’ is Odd, then the Greatest value of nCr is n or nC

3. Permutation

4. Shortcut – 1
Problem based on direct application of the formula, nPr = Solution: If nP4 = 360, find n. nP4 = n(n-1)(n-2)(n-3) = 360 = 6*5*4*3 n = 6.

5. Shortcut – 2
Problems based on formation of numbers with digits when repetition of digits is not allowed.
Working rule:
First of all decide of how many digits the required number will be.
Then fill up the places on which there are restrictions and then apply the formula nPr for filling up the remaining places with remaining digits.

6. Permutation and Combination
Example: How many numbers of four digits can be formed with the digits 1,2,3,4 and 5? (if repetition of digits is not allowed). Solution: Required umber = 5P4 =

7. Shortcut - 3
Number of permutations when repetition is allowed: Working rule: Number of permutations of n different thing taken r at a time when things can be repeated any number of times. = n*n*n…..r times = nr Note: in such type of problems, you have to first determine as to which item can be repeated. And consider the value of repeated item as ‘r’ in the above formula.

8. Permutation and Combination
Example: A gentleman has 6 friends to invite. In how many ways can he send invitation cards to them if he has three servants to carry the cards? Solution: [1st servant, 2nd servant]-- same friend; [1st Friend, 2nd Friend]-- Same servant Here we observe that invitation cards cannot be sent to the same friend by different servants but invitation cards may be sent different friends by the same servant. Thus same servant may be repeated for different friends, therefore, we must start with friend. Invitation cards may be sent to each of the six friends by any one of the three servants in 3 ways. Required Number = 3*3*3*3*3*3 = 36 = 729.

9. Shortcut – 4
If there are two groups A and B consisting of ‘m’ and ‘n’ things respectively, then the number of ways in which no two of group B occur together are given by (m+1Pn*m!) provided that n<m.

10. Permutation and Combination
Example: In a class of 10 students, there are 3 girls. In how many different ways can they be arranged in a row such that no two of the three girls are consecutive? Solution: = m+1Pn*m! = 7+1P3*7! =

11. Shortcut – 5
The number of ways in which ‘n’ examination papers can be arranged so that the best and the worst papers never come together are given by [(n-2)*(n-1)!] ways.

12. Permutation and Combination
Example: In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together. Solution: = [(10-2)*(10-1)!] = 8*9! Note: The number of ways in which ‘n’ books may be arranged on a shelf so that two particular books shall not be together is [(n-2)*(n-1)!]

13. Shortcut – 6
There are ‘m’ boys and ‘n’ girls. The no. of ways in which they can be seated in a row so that all the girls do not sit together are given by [(m+n)! - (m+1)!*n!] ways.

14. Permutation and Combination
Example: There are 5 boys and 3 girls. In how many ways can they be seated in a row so that all the three girls do not sit together? Solution: = (5+3)! – (5+1)! * 3! = 8! – 6! * 3! = 36000

15. Shortcut – 7
The number of ways in which m boys and ‘m’ girls can be seated in a row so that boys and girls alternate are given by 2(m!m!) ways. Solution: = 2(4!4!) = 1152

16. Shortcut – 8
The number of ways in which m boys and (m-1) girls can be seated in row so that they are alternate is given by [m! (m-1)!] Ways.

17. Permutation and Combination
Example: In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate? Solution: = 4!3! = 144.

18. Shortcut – 9
The number of ways in which ‘m’ persons of a particular group, caste, country etc. and ‘m’ persons of the other group, caste, community, country etc can be seated along a circle so that they are alternate, given by [m!(m-1)!] ways.

19. Permutation and Combination
Example: In how many ways can 5 Indian and 5 Englishmen be seated along a circle so that they are alternate? Solution: = 4!5!

20. Shortcut – 10
A round table conference is to be held between ‘n’ delegates. The no. of ways in which they can be seated so that ‘m’ particular delegates always sit together are given by [(n-m)!*m] ways.

21. Permutation and Combination
Example: A round table conference is to be held between 20 countries. In how many ways can they be seated if two particular delegates are always to sit together? Solution: [(20-2)! * 2! = 18! * 2!

22. Permutation and Combination
Note:
Always remember the following results based on the circular permutations.
Number of circular arrangements of ‘n’ different things = (n-1)!
When clockwise and anticlockwise arrangements are not different, number of circular arrangements of ‘n’ different things = ½(n-1)!

23. Shortcut – 11
The number of ways in which ‘n’ different beads can be arranged to form a necklace are given by
If there are n points in a plane and no points are collinear, then the number of straight lines can be drawn using these ‘n’ points are given by

24. Permutation and Combination
Example: Find the number of ways in which 6 different beads can be arranged to form a necklace. Solution: =

25. Shortcut – 12
Find the number of permutations of the letters of the word “Pre-university”. Solution: Total letters = 13, E’s = 2, I’s = 2, R’s = 2. No. of Permutations =

26. Permutation and Combination
How many different words can be formed with the letters of the word “University”, so that all the vowels? Number of letters = 10, No. of vowels = 4, ‘I’ occurred twice. (6+1)! = 7! Required number = 7!

27. Combination

28. Basic Formulae
nCr = n! / [r! (n – r)!] nCr-1+nCr = n+1Cr If nCx = nCy then either x = y or x+y = n

29. Shortcut – 1
If 15C3r = 15Cr+3 , find r. Solution: We know that if nCx = nCy then either x = y or x+y = n 15C3r = 15Cr+3 Either 3r = r +3, which gives r = 3/2 which is not possible, since r is an integer. Or, 3r+r+3 = 15, which gives r = 3.

30. Shortcut – 2
Find the number of ways in which 5 identical balls can be distributed among 10 identical boxes, if not more than one ball can go into a box. Solution: 10C5 =

31. Shortcut – 3
The number of triangles which can be formed by joining the angular points of a polygon of m sides as vertices are
If there are ‘n’ points in a plane and no three points are collinear, then the number of triangles formed with ‘n’ points are given by

32. Permutation and Combination
Example: Find no. of triangles formed by joining the vertices of a polygon of 12 sides. Solution: =

33. Shortcut – 4
The number of diagonals which can be formed by joining the vertices of a polygon of ‘m’ sides are

34. Permutation and Combination
Example: Find the number of diagonals of a polygon of 12 sides. Solution: =

35. Shortcut – 5
If there are ‘m’ horizontal and ‘n’ vertical lines, then the no. of different rectangles formed are given by (mC2*nC2)

36. Permutation and Combination
Example: In a chess board there are 9 vertical and 9 horizontal lines. Find the no. of rectangles formed in the chess board. Solution: = mC2*nC2 = 9C2*9C2 = 36 * 36 = 1296.

37. Shortcut – 6
In a party every person shakes hands with every other persons. If there was a total of H handshakes in the party, the no. of persons ‘n’ who were present in the party can be calculated from the equation given as

38. Permutation and Combination
Example: In a party every person shakes hands with every other persons. If there was a total of 210 handshakes in the party, find the no. of persons who were present in the party. Solution:

39. Shortcut – 7
There are ‘m’ members in a delegation which is to be set abroad. The total no. of members is ‘n’. The no. of ways in which the selection can be made so that a particular ‘r’ members are always.
Included, is given by (n-rCm-r) and
Excluded is given by (n-rCm)

40. Permutation and Combination
There are 5 members in a delegation which is to be sent aboard. The total no. of members is 10. In how many ways can the selection be made so that a particular member is always i) included ii) excluded?
Solution:
Included = 10-1C5-1 = 9C4 = 126
Excluded = 10-1C5 = 9C5 = 126

41. Shortcut – 8
There are ‘n’ points in a plane out of which ‘m’ points are collinear. The number of triangles formed by the points as vertices are given by (nC3 - mC3)
There are ‘n’ points in a plane out of which ‘m’ points are collinear. The number of Straight lines formed by the joining them are given by (nC2 - mC2+1)

42. Permutation and Combination
Example: There are 10 points in a plane out of which 4 are collinear. Find the number of triangles formed by the points of vertices. Solution: = 10C3 - 4C3 = 120 – 4 = 116

43. Shortcut – 9
On a New Year day every students of a class sends a card to every other student. If the postman deliver ‘C’ cards, then the number of students ‘n’ in the class can by the following equation. n(n-1) = C

44. Permutation and Combination
Example: On a New Year day every students of a class sends a card to every other student. The postman delivers 600 cards. How many students are there in the class? Solution: n(n-1) = 600 n2 – n = 0 n = 25, -24 Required number of students = 25.

45. Shortcut – 10
If in an examination a minimum is to be secured in each of ‘n’ subjects for a pass, then the number of ways a student can fail is given by (2n – 1) ways.
If there are ‘n’ questions in a question paper, then the no. of ways in which a student can solve one or more questions are given by (2n – 1) ways.

46. Permutation and Combination
Example: In an examination a minimum is to be secured in each of 5 subjects for a pass. In how many ways can a student fail? Solution: = 25 – 1= 31.

47. Shortcut – 11
In a basket there are certain number of fruits. Out of which, there are ‘x’ oranges, ‘y’ apples, ‘z’ mangoes and the remaining ‘n’ are of different kinds. Then the number of ways a person can make a selection of fruits from among the fruits in the basket are given by [(x+1)(y+1)(z+1)*2n – 1] ways.

48. Permutation and Combination
Example: There are 4 oranges, 5 apples and 6 mangoes in a fruit basket. In how many ways can a person make a selection of fruits from among the fruits in the basket? Solution: [(4+1)(5+1)(6+1)*20 – 1] = 210 – 1 = 209.

49. Shortcut – 12
The number of ways in which (n*m) different things can be divided equally among ‘n’ persons are given by

50. Permutation and Combination
Example: In how many ways 12 different things can be divided equally among 3 persons? Solution: Note: if ‘mn’ different things are divided equally among ‘n’ groups, then the total no of different ways of distribution by

51. Shortcut – 13
The number of ways to distributor or divide ‘n’ identical things among ‘r’ persons when any person may get any number of things are given by [n+r-1Cr-1] ways.

52. Permutation and Combination
Example: In how many ways 20 apples can be divided among 5 boys. = C5-1= 24C4 =

53. Shortcut – 14
The number of Quadrilaterals that can be formed by joining the vertices of a polygon of n sides are given by

54. Permutation and Combination
Example: Find the no. of quadrilaterals that can be formed by joining the vertices of an octagon. Solution:

55. Shortcut – 15
n students appear in an examination. The number of ways the result of the examination can be announced are given by (2)n
n matches are to be played in a chess tournament. The number of ways in which their results can be decided are given by (2)n ways.

56. Shortcut - 15 A badminton tournament consists of ‘n’ matches.
The number of ways in which their results can be forecast are given by (2)n ways.
Total number of forecasts containing all correct results or all wrong results is given by 1.

57. Shortcut – 15
The Indian hockey team is to play ‘n’ matches for the world cup.
The number of different forecasts that will contain all correct results is (1)n or 1.
The number of different forecasts that will contain all wrong results are (2)n.

58. Permutation and Combination
Example: 3 matches are to be played in a class tournament. In how many ways can their results be decided? Solution: = (3)3 = 27.

59. Permutation and Combination
Example:
A badminton tournament consists of 3 matches.
In how many ways can their results be forecast?
How many different forecasts can contain all wrong results?
How many different forecasts containing all correct results or all wrong results are given by 1.
Solution:
Total no. of ways the results of 3 matches can be forecast = 23 = 8
Contain all wrong results = (1)3 = 1
Contain all right results = (1)3 = 1