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Contents
- 1 Permutation and Combination
- 1.1 Permutation
- 1.2 Combination
- 1.3 Real Life Example of Permutation and Combination
- 1.4 Basic(or Fundamental) Principle of Counting
- 1.5 Homework
- 1.6 Factorial Notation
- 1.7 Permutations of Different Objects
- 1.8 Permutations of Objects not all Different
- 1.9 Homework
- 1.10 Permutations of Objects with Repetition
- 1.11 Circular Permutations
- 1.12 Homework
- 1.13 Combinations
- 1.14 Homework
- 2 Binomial Theorem
- 2.1 Binomial Expression
- 2.2 Binomial Theorem
- 2.2.1 Note:
- 2.2.2 Binomial Coefficients
- 2.2.3 General Term
- 2.2.4 Middle Term(s) in the Binomial Expansion
- 2.2.5 Properties of Binomial Coefficients
1 Permutation and Combination
1.1 Permutation
A permutation is an act of arranging numbers or objects in order. It implies all the possible arrangement or rearrangement of the given set, into distinguishable order.Thus, a permutation is an ordered arrangement ı.e. an arrangement in a specific order. For example, all possible permutation created with the letters X, Y, Z
- By taking three all at a time are XYZ, XZY, YXZ, YZX, ZXY, ZYX
- By taking two at a time are XY, YX, YZ, ZY, XZ, ZX
1.2 Combination
The combination is defined as the different ways, of selecting a group, by taking some or all the members of a set, without the following order. ı.e. Combinations are the way of selecting the objects or numbers from a group of objects or collection, in such a way that the order of the objects does not matter. For example,All possible combinations chosen with letter X, Y, Z
- When three out of three letters are to be selected, then the only combination is XYZ.
- When two out of three letters are to be selected, then the possible combinations are XY, YZ, ZX.
Here, XY and YX represent the same combination. However, they are two distinct permuta- tions.
Note:
- When order does not matter, it is a combination.
- When order does matter, it is a permutation.
- Permutation is the ordered combination.
- For a given set of objects, permutation will always be higher than its combination
1.3 Real Life Example of Permutation and Combination
Arranging people, digits, numbers, alphabets, letters, and colours are examples of permutations. Selection of menu, food, clothes, subjects, the team are examples of combinations.
Example-5 How many 3-digit numbers can be formed without using the digits 0, 5 , 6 , 9 when the repe- tition of digit is not allowed?
Solution: Since 0, 5 , 6 , 9 are not to be used, therefore we are left with the digits 1, 2 , 3 , 4 , 7 , 8.ı.e. there are 6 digits Then one’s place can be filled up in 6 ways. Ten’s place can be filled up in 5 ways. Hundred’s place can be filled up in 4 ways. Therefore, by the multiple principle counting, three digit numbers in total are
6 × 5 × 4 = 120
Example-6 How man numbers of atleast three different digits can be formed from the integers 1, 2 , 3 , 4 , 5?
Solution: We have to form a number of atleast three different digits. We can do so by forming a number of three different digits or four different digits or five different digits. We have the digits 1, 2 , 3 , 4 , 5. Then one’s place can be filled up by any one of the 5 digits and so it an be filled up in 5 ways. Since there remains 4 digits to fill up ten’s place, the ten’s place can be filled up in 4 ways. Similarly,the hundred’s place, the thousand’s place and the ten thousand’s place can be filled up in 3 ways, 2 ways and 1 way respectively.
To form a number of three different digits, we have to fill up one’s, ten’s and hundred’s places.Therefore, by the multiplication principle of counting a number of three digits can be formed in 5 × 4 × 3 = 60 ways
To form a number of four different digits, we have to fill up one’s, ten’s, hundred’s and thousand’s places. Therefore, by the multiplication principle of counting, a number of four different digits can be formed in
5 × 4 × 3 × 2 = 120 ways
To form a number of five different digits, we have to fill up the one’s, ten’s, hundred’s, thousand’s and ten thousand’s places. Therefore by multiplication principle of counting, a number of five different digits can be formed in
5 × 4 × 3 × 2 × 1 = 120 ways
Therefore, a number of atleast three different digits can be formed in
60 + 120 + 120 = 300 ways
1.5 Homework
- In a hall, there are three entrance doors and two exit doors. In how many ways can a person enter the hall then come out? Ans : 6 ways
- In a school, there are 8 doors, in how may ways can a student enter the school and leave by different door? Ans : 56 ways
- In how many ways can 4 people be seated in a row containing 6 seats? Ans : 360 ways
- A football stadium has 4 entrance gates and 9 exits. In how many different ways can a man enter and leave the stadium? Ans : 36 ways
- There are 6 doors in a hostel.In how many ways can a student enter the hostel and leave by a different door? Ans : 30 ways
- In how many ways can a man send 3 of his children to 7 different college of a certain town? Ans :
- Suppose there are 5 main roads between the cities A and B. In how many ways can a man go from a city to the other and return by different? Ans : 20
- How many numbers of atleast 3 different digits can be formed from the integers 1, 2 , 3 , 4 , 5 , 6? Ans : 1920
- How many numbers of at most 3- different digits can be formed the integers 1, 2 , 3 , 4 , 5? Ans: 85
- How many numbers are there between 200 and 900 in which all the digits are distinct? Ans: 504
- How many ways 3-digit odd numbers can be formed by using the digit 1 to 9 when the repetition of digit is not allowed? Ans : 280
- Telephone numbers consists of 7 digits and none of them begin with zero. How many tele- phone numbers could be possible if no digit appears more than once? Ans: 544320
- How many numbers between 4000 and 5000 can be formed with the digits 2, 3 , 4 , 5 , 6 , 7? Ans : 60
- How many numbers of 3 digits can be formed from the integers 2, 3 , 4 , 5 , 6? How many of them divisible by 5? Ans : 60, 12
After filling up the second position, there remain n − 2 objects and so the third position can be filled up in n − 2 ways. [n − 2 = n − (3 − 1)] In this way continue to fill up the positions successively. rth^ position can be filled up in n−(r−1) = n − r + 1 ways. Then by the multiplication principle of counting, all the r positions can be filled up one after another in n(n − 1)(n − 2) · · · (n − r + 1) ways. Therefore, the number of permutations of n distinct objects taken r objects at a time is given by P (n, r) = n(n − 1)(n − 2) · · · (n − r + 1)
=
n(n − 1)(n − 2) · · · (n − r + 1) × (n − r) × (n − r − 1) × · · · × 3 × 2 × 1 (n − r) × (n − r − 1) × · · · × 3 × 2 × 1
=
n! (n − r)!
Remark: when r = n, nPn = P (n, n) = n! (by above theorem) Thus, the number of permutation of n distinct objects taken all at a time is P (n, n) = n!
Example-1 If 3 persons enter a bus in which there are 10 vacant seats, in how many ways can they take their seats?
Solution: No. of vacant seats, n = 10 No. of persons, r = 3 Total number of permutations, P (n, r) =?
we have, P (n, r) =
n! (n − r)!
⇒ P (10, 3) =
10 × 9 × 8 × 7!
Therefore, the total number of ways in which they can take their seats is 720
Example-2 Find the number of permutations of the three letters X, Y, Z taken all at a time.
Solution: The number of permutations of 3 letters taken all at a time is
P (3, 3) =
= 3 × 2 × 1 = 6
Example-3 8 boys are participating in a race. How many ways can the first 3 prizes be won?
Solution: There are 8 boys only 3 of them may win the first 3 prizes. Here, n = 8, r = 3
we have, P (n, r) =
n! (n − r)!
⇒ P (8, 3) =
8 × 7 × 6 × 5!
= 8 × 7 × 6 = 336
∴ The first three prizes may be won in 336 ways.
Example-4 In how many ways can 5 boys and 4 girls be arranged for a group photograph if the girls are to seat on chairs in a row and the boys are to stand behind them?
Solution: There are 4 chairs and 4 girls. So, the 4 girls can seat in P (4, 4) ways. Since 5 boys stand behind the girls, so they can stand in P (5, 5) ways. Then by multiplication principle of counting, the girls and boys can be arranged in
P (4, 4) × P (5, 5) = 4! × 5! = 4 × 3 × 2 × 1 × 5 × 4 × 3 × 2 × 1 = 2880 ways
Example-5 It is required to seat 5 boys and 4 girls in a row so that the girls occupy the even places. How many such arrangements are possible?
Solution: Suppose the seats in the row are numbered from 1 to 9. Then the girls can occupy 2nd, 4 th, 6 th, 8 th seats ı.e. there are 4 seats for 4 girls Thus, 4 girls can sit in P (4, 4) ways. The 5 boys can sit in the remaining 5 seats in P (5, 5) ways. Therefore, by the multiplication principle of counting, the girls and boys can be arranged in
P (4, 4) × P (5, 5) = 5! × 4! = 2880 ways
Example-6 Find the number of ways in which 5 boys and 4 girls can be seated in a row so that no two girls are together.
Solution: The 5 boys can occupy their seats in P (5, 5) ways. Now for the girls there are 6 seats ( 4 seats in between boys, 1 seat before the first boy and 1 seat behind the last boy). Thus, the 4 girls can be arranged in P (6, 4) ways. Thus, by the multiplication principle of counting, the boys and girls can be seated in a row so that no two girls are together in
P (5, 5) × P (6, 4) = 5! ×
= 5 × 4 × 3 × 2 × 1 ×
6 × 5 × 4 × 3 × 2!
= 43200 ways
Example-7 In how many ways can eight people can be seated in a row of eight seats so that two particular persons are always together?
Solution: 2 particular persons who wish to seat together can be consider to be one. Then there will be 7 person and 7 seats. The 7 persons can be arrange in P (7, 7) ways. On the other hand, the 2 persons sitting together can interchange their seats in 2! = 2 ways. Therefore,by the multiplication principle of counting, all the persons can be arranged in
P (7, 7) × 2 = 7! × 2 = 7 × 6 × 5 × 4 × 3 × 2 × 1 × 2 = 10080 ways
The required number of permutations is given by
P =
n! p!q!r!
=
11 × 10 × 9 × 8 × 7 × 6 × 5 × 4!
4 × 3 × 2 × 2 × 1 × 2 × 1 × 4!
Example-2 In how many ways can the letters of the word ′CALCU LU S′^ be arranged so that the two C’s do not come together?
Solution: The word ′CALCU LU S′^ contains 8 letters among them there are 2 C’s, 2 U’s and 2 L’s. Thus, the total no. of different arrangements
=
8 × 7 × 6 × 5 × 4 × 3 × 2!
2 × 1 × 2 × 1 × 2!
Let 2 C’s come together. Then considering 2 C’s to be one, there will be 7 letters in which 2 U’s and 2 L’s. Thus, if two C’s come together, then the total number of different arrangements
=
7 × 6 × 5 × 4 × 3 × 2!
2! × 2 × 1
Hence, the number of arrangements in which the 2 C’s do not come together is
5040 − 1260 = 3780
Example-3 How many words can be formed from the letters of the word ′EN GLISH′? How many of these do not begin with E? How many of these begin with E and do not end with H?
Solution: There are 7 letters in the word ′EN GLISH′^ and all letters are different. Therefore, they can be arranged in P (7, 7) = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040 ways Second Part: We see that the first position can be filled up by any one of the letters N, G, L, I, S, H. So, it can be filled up by 6 ways. The remaining position can be filled up by any of the remaining letters including E. Therefore, they can be filled up in P (6, 6) ways. ∴ By the multiple principle of counting, the required number of words that do not begin with E is 6 × P (6, 6) = 6 × 6! = 6 × 6 × 5 × 4 × 3 × 2 × 1 = 4320 ways. Last Part: The first place can be filled up by the letter E. ı.e. it can be filled up by 1 way. Since each arrangements must not end with H. So, last position can be filled up by the letters N, G, L, I, S ı.e. , it can be filled up in 5 ways. Also, the remaining 5 positions can be filled up by any of the remaining 5 letters including H. Thus, they can be filled up in P (5, 5) ways. Therefore, by the multiple principle of counting, the required number of permutations of the letters beginning with E and not ending with H is 1 × 5 × P (5, 5) = 5 × 5! = 5 × 5 × 4 × 3 × 2 × 1 = 600 ways.
Example-4 In how many ways can the letters of the word ′N EP AL′^ be arranged so that
a) all the vowels are always together? b) the vowels may occupy odd positions? c) the relative positions of vowels and consonants are not changed?
Solution: a) There are two vowels. If two vowels come together, consider two vowels as one vowel.Then there are 4 letters,they can be arranged in P (4, 4) ways.Also two vowels position can be interchange in 2! = 2 ways. Therefore, by the multiple principle of counting, number of arrangements when the vowels are together is
2 × P (4, 4) = 2 × 4! = 2 × 4 × 3 × 2 × 1 = 48
b) There are three odd positions for the 2 vowels. So, they can be arranged in P (3, 2) ways. Also, the remaining three letters can be arranged in P (3, 3) ways. Therefore,by multiplication principle of counting, the total number of arrangements
P (3, 2) × P (3, 3) =
× 3! =
3 × 2 × 1
× 3 × 2 × 1 = 6 × 6 = 36 ways
c) Two vowels can be arranged in P (2, 2) = 2! ways and 3 consonants can be arranged in P (3, 3) = 3! ways. Therefore, total number of arrangements = 2! × 3! = 12 ways.
1.10 Permutations of Objects with Repetition
Theorem 1.3. The number of permutations of n objects taken r at a time with repetition is nr. Proof. Suppose we have to fill r position with n different objects and each object can be used as often as we please. Since any of n objects can be filled up in the first position, the first place can be filled in n ways. After the first position filled in, we will have again n objects to fill up second position, because the object which occupies the first position can also be placed in the second position. Hence the second position can also be filled in n ways. Similarly, each of the remaining positions can be filled in n ways. Therefore, by the multiplication principle of counting the r positions can be filled in n × n × n × · · · × n ︸ ︷︷ ︸ r times
= nr
Example-1 In how many ways 3 books be distributed among 4 boys, each boy is eligible for all books.
Solution: Each of 3 books can be distributed in 4 in ways. Hence the required number of ways = 4 × 4 × 4 = 4^3 = 64
Example-2 A boy has 6 pockets. How many ways can be put 5 marbles in his pocket?
Solution: Given that number of pockets = 6 and number of marbles = 5 There is no restriction to put the marbles. Therefore, 5 marbles can be put in 6 × 6 × 6 × 6 × 6 = 6^5 = 7776
1.11 Circular Permutations
The permutations that we have considered so far are called linear permutations because they are the arrangements of the objects in a line or row. Since a line has two end-points, in linear permutations we can name the objects; first, second, third and so on. We cannot do so in case of circular arrangements, as there is neither beginning nor an end in circular arrangement, ı.e. no end points. However, we can consider the position of each object in a circular arrangement relative to other objects in the arrangement. Hence while arranging objects in a circle, one object is kept fixed and the remaining objects are then arranged relative to it. In such arrangement, we have to consider relative positions of the different objects, ı.e., the circular permutations are different only when the relative order of the objects is changed, otherwise they are same. For example, The arrangements BCDA, CDAB, DABC are same as the arrangement ABCD. Theorem 1.4. The number of circular permutations of n different objects taken all at a time is (n − 1)!
Proof. In circular arrangement the position of each object is determined relative to other objects in the arrangement. Hence the first object can be arranged in only one way, because for the first object there will not be the question: relative to which the first object is to be kept fixed. After arranging, the first object, the remaining n − 1 objects are arranged relative to the first and so, they can be arranged in (n − 1)! ways. Then by multiplication principle of counting the first object followed by the remaining objects can be arranged in
1 × (n − 1)! = (n − 1)! ways.
∴ The number of permutations of n objects taken all at a time is (n − 1)!
Alternatively,
Proof. Let a 1 , a 2 , a 3 , · · · , an− 1 , an be n distinct items. Let the total number of circular permutations be x. There are n! ways in which they can be arranged in a row. On the other hand, all the linear arrangements
a 1 a 2 a 3 · · · an− 1 an a 2 a 3 a 4 · · · ana 1 a 3 a 4 a 5 · · · ana 1 a 2 · · · · · · ana 1 a 2 · · · an− 1
will lead to the same arrangements for a circular table. Hence one circular arrangement corre- sponds to n unique row (linear) arrangements. But there are x circular permutations. So, total number of linear permutations is x × n. Since the number of linear permutations of n distinct items is n!.
Therefore, nx = n!
or, x =
n! n = (n − 1)!
Remark: In above theorem anticlockwise and clockwise order of arrangements are considered as distinct permutations.
Note:
- When the positions are numbered, circular arrangements is treated as a linear arrangement.
- In linear arrangements it does not make difference whether the positions are numbered or not.
- Number of circular permutations of n different things taken r at a time is nPr r
P (n, r) r
[As each circular permutation gives r linear permutations, x × r = P (n, r)]
Example-5 suppose the prime ministers of 7 South countries meet together to discuss a problem. In how many ways can they sit together at a round table if India and Pakistan prime ministers are not to sit together?
Solution: The prime minister of 7 countries can be arranged at a round table in (7 − 1)! = 6! = 720 ways. If the prime ministers of India and Pakistan were to sit together, consider two prime ministers as one then there will be 6 prime ministers in all and they are arranged in a round table in (6 − 1)! = 5! = 120 ways. But the prime ministers of India and Pakistan could be seated in 2! = 2 ways. Thus, if India and Pakistan Prime ministers were sit together, then total no. of arrangements = 2 × 120 = 240 Therefore, if India and Pakistan prime ministers are not sit together,then the required number of arrangements = 720 − 240 = 480
Example-6 If 6 persons were invited for a party, in how many ways can they and the host be seated at a circular table? In how many ways they can be seated if two particular persons must be seated on either side of the host?
Solution: There are 7 persons with host. 7 persons can be seated in a circular table in (7 − 1)! = 6! = 720 ways Second part: If host and two particular persons sitting on either side of the host, consider 3 as a one. Then there will be 5 persons and they can be arranged at a circular table in (5 − 1)! = 4! = 24 ways. But two particular persons sitting on either side of the host can take their seats in 2! = 2 ways Therefore, the number of ways of arranging 7 persons at a circular table with 2 particular persons on either side of the host is
2 × 24 = 48
1.12 Homework
- In how many ways can 4 Arts students and 4 science students be arranged alternately at a round table? Ans: 144
- 6 girls form a round for dancing. In how many ways they stand in the table? Ans: 120
- Find the number of ways in which 10 different flowers can be strung to form a garland so that 4 particular flowers are always together. Ans: 8640
- A round table meeting is to be held between the delegates of 9 countries. In how many ways can they be seated if two particular delegates may wish to together? Ans: 10080
- In how many ways can 9 members of a committee sit at a round table so that secretary and the joint secretary are always the neighbours of the president? Ans:
- In how many ways can a conference of 5 boys and 5 girl be seated at a circular table so that no two girls are adjacent?
- In how many different ways can a garland of 9 flowers be made? Ans: 20160
- In how many different ways can 10 different pearls be arranged to form a necklace? Ans: 181440
- In how many ways can 3 letters be posted in 4 letter boxes? Ans:
- In how many ways can seven different coloured beads be made into a bracelet? Ans:
- In how many ways can 4 letters be posted in 6 letter boxes? Ans:
- In how many ways can 3 prizes be distributed among 4 students so that each student may receive any number of prizes? Ans:
- How even numbers of three digits can be formed when the repetition of digits is allowed? Ans:
Corollary 1.5.1. The number of combination of n distinct items taken r at a time is equal to the number of combination of n distinct items taken (n − r) items at a time. That is,
C(n, r) = C(n, n − r)
Proof.
C(n, n − r) =
n! (n − n + r)!(n − r)!
=
n! r!(n − r)! ∴ C(n, n − r) = C(n, r)
Corollary 1.5.2. If C(n, r) = C(n, s) then either r = s or r + s = n
Proof. Given that
C(n, r) = C(n, s) ∴ r = s
Again given,
C(n, r) = C(n, s) If r 6 = s, then C(n, s) = C(n, r) = C(n, n − r) [using above corollary] ⇒ s = n − r ⇒ r + s = n
∴ either r = s or r + s = n
Corollary 1.5.3. C(n, r) + C(n, r − 1) = C(n + 1, r) (1 ≤ r ≤ n)
Proof.
C(n, r) + C(n, r − 1) =
n! (n − r)!r!
n! (n − r + 1)!(r − 1)!
=
n! (n − r)! × r × (r − 1)!
n! (n − r + 1) × (n − r)! × (r − 1)!
=
n! (n − r)!(r − 1)!
×
r
n − r + 1
n! (n − r)!(r − 1)!
×
n − r + 1 + r r(n − r + 1)
(n + 1)n! (n − r + 1)(n − r)! × r(r − 1)!
=
(n + 1)! (n + 1 − r)!r! = C(n + 1, r) ∴ C(n, r) + C(n, r − 1) = C(n + 1, r)
Examples:
Example-1 If P (n, r) = 720 and C(n, r) = 120, find r
Solution: We know that
C(n, r) =
P (n, r) r! ⇒ r! =
P (n, r) C(n, r)
=
⇒ r! = 6 = 3 × 2 × 1 = 3! ∴ r = 3
Example-2 From a class of 20 students 4 are to be chosen for a competition. In how many ways can this be done?
Solution: We have to select 4 students out of 20 students. Here, n = 20, r = 4
C(n, r) =
n! (n − r)!r!
20 × 19 × 18 × 17 × 16!
16! × 4 × 3 × 2 × 1
Therefore, 4 students can be selected out of 20 students in 4845 ways.
Example-3 If there are 10 persons and if each two of them shakes hands with each other, how many hand shakes happen in the party?
Solution: Each two of them shakes hand with each other. In other words, we have to select 2 persons out of 10 persons.Setting, n = 10 r = 2, we get
C(n, r) =
n! (n − r)!r!
10 × 9 × 8!
8! × 2 × 1
Therefore, 45 handshakes happen in the party.
Example-4 In how many ways can a student choose 5 courses out of 9 courses, if 2 courses are compulsory for every student?
Solution: A student has to choose 5 courses out of 9 courses, but 2 courses are compulsory for every student. That means, the student is allowed to choose 3 courses out of 7 courses.
Now C(7, 3) =
7 × 6 × 5 × 4!
4! × 3 × 2 × 1
Therefore, the student can choose 5 courses out of 9 courses in 35 ways, if 2 course are compulsory for each student.
