This document discusses fundamental principles of counting and permutation and combination. It provides examples to illustrate key concepts such as: 1) Mohan has 3 pants and 2 shirts - there are 3×2 = 6 ways to choose a pant and shirt combination. 2) Sabnam has 2 bags, 3 boxes, and 2 bottles - the number of ways to choose one of each is the number of ways to successively choose a bag, box, and bottle. 3) With 4 roads between a student's house and college, there are 4×3 = 12 ways for the student to go to college and return on different roads.

1. Permutation & Combination
Dr. N. B. Vyas
Dept. of Humanities & Science
Atmiya University

2. Fundamental Principle of Counting
• Let us consider the following problem.
• Mohan has 3 pants and 2 shirts. How many different
pairs of a pant and a shirt, can he dress up with?
• There are 3 ways in which a pant can be chosen,
because there are 3 pants available.
• Similarly, a shirt can be chosen in 2 ways. For every
choice of a pant, there are 2 choices of a shirt.
• Therefore, there are 3 × 2 = 6 pairs of a pant and a
shirt.

3. • Let us name the three pants as P1, P2 , P3 and
the two shirts as S1, S2.
• Then, these six possibilities can be illustrated in
the Figure.

4. • Let us consider another problem of the same
type.
• Sabnam has 2 school bags, 3 tiffin boxes and 2
water bottles.
• In how many ways can she carry these items
(choosing one each).

6. • In the first problem, the required number of
ways of wearing a pant and a shirt was the
number of different ways of the occurrence of
the following events in succession:
(i) the event of choosing a pant
(ii) the event of choosing a shirt.
• In the second problem, the required number of
ways was the number of different ways of the
occurrence of the following events in succession:
(i) the event of choosing a school bag
(ii) the event of choosing a tiffin box
(iii) the event of choosing a water bottle.

7. Example
• A student wants to go to his college from his
house. There are four different roads between
his house and the college. He wants to go to the
college by one road and does not want to return
by the same road. In how many different ways
can he go to the college and come back?
• Let us denote the four roads by A, B, C and D.
He can go to college and return in the following
different ways:

8. 1. He goes by road A and return by road B.
2. He goes by road A and return by road C.
3. He goes by road A and return by road D.
4. He goes by road B and return by road A.
5. He goes by road B and return by road C.
6. He goes by road B and return by road D.
7. He goes by road C and return by road A.
8. He goes by road C and return by road B.
9. He goes by road C and return by road D.
10.He goes by road D and return by road A.
11. He goes by road D and return by road B.
12. He goes by road D and return by road C.

9. • Thus there are 12 different ways of going to the
college and coming back.
• The problem can be thought in different way:
• There are four different alternatives of his going
to the college, and with each alternative of going
to the college, there are (remaining) three
different alternatives of returning from the
college.
• Hence there are 4 x 3 = 12 different ways in
which he can go to the college and come back.

10. Example
• Let us consider there are three senior professors
in a college. Out of them one is to be appointed
as principal and another as vice principal.
• In how many different ways this can be done?
• Let us denote the three professors as A, B and C.
The appointments can be made in the following
different ways:

11. 1. A as principal and B as vice principal.
2. A as principal and C as vice principal.
3. B as principal and A as vice principal.
4. B as principal and C as vice principal.
5. C as principal and A as vice principal.
6. C as principal and B as vice principal.
• Thus appointments can be made in 6 different
ways as shown above.

12. • Here any once from A, B and C can be appointed
as principal. Hence appointment of principal can
be made in 3 different ways.
• Once the principal is appointed, any one from
remaining 2 professors can be appointed as vice
principal. Thus for each way of appointing a
principal, there are two ways of appointing a
vice-principal.
• Hence appointments of principal and vice-
principal can be done in 3 x 2 = 6 ways

13. Example
• Suppose we want to arrange two things out of
five things A, B, C, D and E.
• The different arrangements of two things out of
five things can be made as follows:

14. Fundamental principle of counting OR
The multiplication principle
• If an event can occur in m different ways,
following which another event can occur in n
different ways, then the total number of
occurrence of the events in the given order is
m×n.

15. Example 1
• Find the number of 4 letter words, with or
without meaning, which can be formed out of
the letters of the word ROSE, where the
repetition of the letters is not allowed.

16. Solution
• There are as many words as there are ways of
filling in 4 vacant places _ _ _ _
• by the 4 letters, keeping in mind that the
repetition is not allowed.
• The first place can be filled in 4 different ways by
anyone of the 4 letters R,O,S,E.
• Following which, the second place can be filled
in by anyone of the remaining 3 letters in 3
different ways,
• following which the third place can be filled in 2
different ways; following which, the fourth place
can be filled in 1 way.

17. • Thus, the number of ways in which the
• 4 places can be filled, by the multiplication
principle, is 4 × 3 × 2 × 1 = 24.
• Hence, the required number of words is 24.

18. Example 2
• Given 4 flags of different colours, how many
different signals can be generated, if a signal
requires the use of 2 flags one below the other?

19. Permutation
• nPr = n (n-1) (n-2) (n-3) …. (n-r+1)
• nPr = n! / (n-r)!
• nPn=n!

20. Q.1 Find the values
1. 9P4
2. 7P5
3. 20P3
4. 6P6
5. 50P2

21. Q.2 Find the value of n
1. nP3 = 210
2. 6Pr = 120
3. 210 n! = 7!
4. 3. nP3 = 2 (n+1)P3

22. Q.3
• How many different numbers of four digits can
be formed from the digits 2, 3, 5, 6 and 8 ? Each
digit should be used only once in each number.

23. Solution
• We have five digits 2, 3, 5, 6 and 8 and out of
them four digits are to be arranged.
• Therefore,
• Total number of permutation
• = 5P4
• = 5 x 4 x 3 x 2 x 1
• = 120
• Thus 120 different numbers of four digits can be
formed.

24. Q.4
• How many different numbers of five digits can
be formed by using all digits of number 50231 ?

25. Solution
• In forming a five digit number, the 1st place can
be filled up from any of the four digits 5, 2, 3, 1
because first place cannot be filled by 0.
• So first place can be filled up in 4 ways.
• 0 can be placed in 2nd place so second place can
be filled up in 4 ways.
• 3rd place in 3 ways, 4th place in 2 ways and fifth
place in 1 way. Therefore,
• Total number of ways = 4 x 4 x 3 x 2 x 1 =

26. Q.5
• How many new words can be formed by using all
the letters of the word LOGARITHM ?
• In how many of these words L will be at the first
place?
• In how many words L will be at the first place
and M will be at the last place?

27. Solution:
• There are 9 letters in word LOGARITHM. The
total number of permutations of these 9 letters.
• = 9P9 = 9! =_____
• 362880
• As LOGARITHM is one of the words among
them, the total number of new words
• = 362880 – 1 = ______
• If L is to be kept at the first place it can be
arranged in 1 way and the remaining 8 letters
can be arranged in 8! Ways.

28. • Therefore, total permutations in which L is at
the first place
• = 1 x 8! = ______
• 40320
• New words in which L at the first place
• = 40320 – 1 =_____
• If L is to be kept at the first place and M is to be
kept at the last place, the remaining 7 letters can
be arranged in 7! Ways.
• Therefore, total number of permutations in
which L is at the first place and M is at the last
place
• = 1 x 1 x 7! = _____
• New words in which L is at the first place and M
is athe last place = 5040 – 1 = _____

29. Q.6
• 4 books of statistics, 3 books of Economics and 5
books of Accountancy are to be arranged on a
shelf in one row. Find the number of ways in
which the arrangements can be made if the
books of the same subject are to be kept
together.

30. Solution:
• 4 books of statistics can be arranged in 4! ways,
• 3 books of Economics can be arranged in 3!
ways,
• 5 books of Accountancy can be arranged in 5!
ways
• Moreover the three subjects can be permuted in
3! ways.
• Therefore, total number of permuations
• = 4! x 3! x 5! X 3! = ______

31. Q.7
• How many different words can be formed from
the letters of the word GASOLINE, so that
consonants always occupy the odd places and
vowels occupy the even places.

32. Solution
• In the word GASOLINE, there are four
consonants G, S, L, N and four vowels A, O, I, E.
• The four consonants can be arranged at odd
places in 4! Ways
• And the four vowels can be arranged at even
places in 4! ways
• Therefore total number of permutations
• =4! x 4! = _______

33. Q.8
• How many different numbers of four digits
divisible by 5 can be formed from the digits 2, 3,
5, 7, 8?

34. Solution
• In order that number is divisible by 5, the last
digit should be 5.
• This can be arranged in one way.
• The first three places can be filled up from the
remaining four digits 4P3 ways.
• Therefore total number of permutations:
• = 1 x 4P3
• = 1 x 24 = ______

35. Q.9
• Four boys and four girls are to be arranged in a
row in such a way that no two boys or two girls
are side by side. Find the number of ways in
which the arrangements can be made.

36. Solution
• Denoting girl by G and boy by B, the
arrangements can be made in the following ways
• G B G B G B G B
• B G B G B G B G
• In the first arrangement the girls are in 1st, 3rd ,
5th and 7th places. The girls can be permuted in
4P4 ways.
• Similarly the boys can be permutted in 2nd, 4th ,
6th and 8th places in 4P4 ways.

37. • Therefore, total number of permutations in
which girl is at the first place
• = 4P4 x 4P4
• =24 x 24 = ______
• =576
• Similarly the total number of permutations in
which boy is at the first place = ________
• =576
• Thus the total number of permutations in which
no two boy or no two girl are side by side
• = 576 +576 = _______

38. Q.10
• Four professors, four boys and two girls are to be
seated in chairs in a row for a photograph. If
boys are to sit on two chairs at the end of either
side and the girls are not to sit with the boys,
find the number of ways in which arrangements
can be made.

39. Solution
• There are ten chairs in all:
• Two chairs on each end are to be occupied by the
boys. i.e. chair number 1,2,9,10 are to be occupied by
boys.
• 4 boys can be arranged in 4 chairs in 4P4 = 4! ways
• Girls are not allowed to sit in chair number 3 and 8.
hence in the chair number 4,5,6,7 two girls can sit in
4P2 ways.
1 2 3 4 5 6 7 8 9 10

40. • Four professors can occupy the remaining four
chairs in 4P4 = 4! Ways
• Therefore total number of arrangements
• = 4! x 4P2 x 4!
• =______
• =6912

41. Permutations of Similar things
• The number of permutations of n things of
which p things are of one kind, q things are of
second kind, r things are of third kind and
remaining are of different kinds is given by
!
! ! !
n
p q r

42. Ex.1
• Find the number of permutations of all letters of
word “BOOK”.

43. Solution:
• In the word “BOOK” there are four letters in
which ‘O’ occurs twice.
• Therefore total number of permutations
• = 4! / 2!
• =____
• =12

44. Ex.2
• How many different words can be formed by
using letters of the word “MISSISSIPPI”?

45. Solution:
• Here in the word “MISSISSIPPI” there are 11
letters in which ‘S’ occurs 4 times, ‘I’ occurs 4
times and ‘P’ occurs 2 times.
• Therefore total number of permutations
• = 11! / (4! 4! 2! )
• =_______
• =34650

46. Permutations when things are
repeated:
• If two things are to be arranged out of four
things A, B, C, D and if things elected for the first
time can also be selected for the second time
then the number of arrangements are as follows:
• AA AB AC AD
• BA BB BC BD
• CA CB CC CD
• DA DB DC DD
• The total number of permutations is 16.

47. • In this case, the 1st thing can be arranged in 4
ways and similarly 2nd , 3rd and 4th things can
also be arranged in 4 ways.
• Therefore total permutations:
• = 4 x 4
• =42
• =_____

48. Ex.3
• How many different numbers of three digits can
be formed from the digits 2, 3, 4, 5, 6. Each digit
can be occur any number of times.

49. Solution:
• There are 5 digits i.e 2, 3, 4, 5, 6
• The total number of different numbers of three
digits when the repetitions are allowed
• = 53
• =5 x 5 x 5
• =_____

50. Ex.4
• Find the total number of permutations when
three cubical dice are thrown

51. Solution:
• There are 3 dice.
• Each dice are having numbers 1, 2, 3, 4, 5, 6 on it
• If all 3 dice are thrown, the total number of
permutations = 63
• =____
• =216

52. Restricted permutations
• The number of permutation of n different things
taken r at a time in which x particular things do
not occur is (n-x)Pr

53. Ex.1
• How many four digit numbers which are
divisible by 5 can be formed with 0, 1, 2, 3, 5, 7, 9
such that no digit is repeated?

54. Solution:
• In order that the number is divisible by 5, the
last digit must be 0 or 5. we are given 7 digits i.e.
0, 1, 2, 3, 5, 7, 9
• In 4 digit number if the last place is given to 0
then the remaining first 3 places can be filled up
in 6P3 ways = _____
• =120 ways
• If the last place is given to 5, the first place
cannot be given to 0, and hence the number of

55. • Arrangements = 1P1 x 5P1 x 5P2
• =1 x 5 x 20 =_______
• = 100
• Therefore total number of permutations
• =120 + 100 = _______

56. Ex.2
• Find r, if 7Pr = 60 x 7P(r-3)

57. Ex.3
• Find k, if 8P5 = 7P5 + k 7P4

58. Ex.4
• Prove that: nPr =n . (n-1)P(r-1)

59. Ex.5
• How many words can be formed by using the
letters of the word “BUSINESS”?
• Find the number of words in which all the three
‘S’ are together.
• Also find the number of words which start with
‘BUSI’

60. Solution:
• “BUSINESS” consists of 8 letters including three
‘S’
• Therefore total number of words that can be
formed using the letters of the word
“BUSINESS”
• = 8! / 3! =_____
• =6720

61. • If three ‘S’ are to be kept together, they can be
regarded as one unit. Hence we have 1 group + 5
other letters = 6 in all.
• They can be arranged in 6P6 = ____
• =720 ways
• For the word starting with ‘BUSI’. Fixing ‘BUSI’,
the remaining 4 places can be filled up by N, E,
S, S.
• Total arrangements = 4! / 2!
• = 12 ways

63. • 1. How many 3-digit numbers can be formed from the
digits 1, 2, 3, 4 and 5
• assuming that
• (i) repetition of the digits is allowed?
• (ii) repetition of the digits is not allowed?
• 2. How many 3-digit even numbers can be formed from the
digits 1, 2, 3, 4, 5, 6 if the
• digits can be repeated?
• 3. How many 4-letter code can be formed using the first 10
letters of the English
• alphabet, if no letter can be repeated?
• 4. How many 5-digit telephone numbers can be
constructed using the digits 0 to 9 if
• each number starts with 67 and no digit appears more than once?
• 5. A coin is tossed 3 times and the outcomes are recorded.
How many possible
• outcomes are there?
• 6. Given 5 flags of different colours, how many different
signals can be generated if
• each signal requires the use of 2 flags, one below the other?