Addition principle:

If one experiment has n possible outcomes and another has m possible outcomes, then there are (m + n) possible outcomes when exactly one of these experiments is performed. In other words, if a job can be done by n methods and by using the first method, can be done in a₁ ways or by second method in a₂ ways and so on . . . by the nth method in a_n ways, then the number of ways to get the job done is (a₁ + a + ... + a_n).

Example.1

A college offers 7 courses in the morning and 5 in the evening. Find the number of ways a student can select exactly one course, either in the morning or in the evening.

Solution

The student has seven choices from the morning courses out of which he can select one course in 7 ways.

For the evening course, he has 5 choices out of which he can select one course in 5 ways. .

Hence he has total number of 7 + 5 = 12 choices.

Example.2

A person wants to leave station B. There are three routes from station B to A and four routes from B to C. In how many ways can he leave the station B.

Solution

B → A in 3 ways

B → C in 4 ways

&⇒ He can leave station B in 3 + 4 = 7 ways.

Example.3

How many three digit numbers xyz with x, z < y can be formed

Solution

Obviously 2 £ y £ 9. If y = k, then x can take values from 1 to k – 1 and z can take values from 0 to k-1. .

Thus required number of numbers = .

Bijection Principle :

Generally it is not easy to count the numbers belonging to a very large collection(set) directly. Let us assume that we have a set X and we want to count n(x) ( number of elements in X). And we have another set Y and n(Y) can be counted(easily). Now if there exists one-one and onto relationship from set X to set Y, then n(X) = n(Y) . .

In other words, if X and Y are two finite sets such that there is a bijection ( i.e. one-one onto map) from X to Y , then n(X) = n(Y). .

For example let us take a simple case. The sitting capacity of a class is 30. On a particular day, the class is full then, without actual counting, we can say that the total number of the students present in the class is 30. .

Inclusion-Exclusion Principle:

Let U be a finite set of m elements and p₁ , p₂ , p₃ , . . . ,p_r be some properties which the objects of U may or may not have, .

Then the number of elements of U which have at least one of the properties p₁, p₂, p₃, …. ,p_r is given by .

= S₁ – S₂ + S₃ – S₄ + . . . +(–1)^r-1S_r .

and the number of elements of U which have none of the properties p₁, p₂, p₃ , . . . ,p_r is given by = m – S₁ + S₂– S₃ + …. + (– 1) ^r S_r .

Where A_i denotes the subset of U each of whose elements has the property p_i( i = 1, 2, …. , r) and denotes the subset of U none of whose elements has the property p_i

A_i Ç A_j denotes the set of all elements of U having both properties p_i and p_j and so on. And S_k denotes the total number of elements of U having any k of the given r properties. .

E.g. S₁ = n(A₁) + n(A₂) + …. + n(A_r) , S₂ = n(A₁ÇA₂) + n(A₁Ç A₃) +…. + n(A₁ÇA_r) + n(A₂ÇA₃) + . . . + n(A₂ÇA_r) + …. + n(A_r-1 Ç A_r) . .

In particular case:

For r =2 , n(A₁ÈA₂) = n(A₁) +n(A₂) – n(A₁ÇA₂)

For r = 3,

n(A₁ÈA₂ ÈA₃) = n(A₁) +n(A₂) +n(A₃) – n(A₁ÇA₂) – n(A₁ÇA₃) – n(A₂ÇA₃) + n(A₁ Ç A₂ ÇA₃). .

For r = 4, n(A₁ÈA₂ÈA₃ÈA₄) = n(A₁) +n(A₂) +n(A₃) + n(A₄) – n(A₁ÇA₂) – n(A₁ÇA₃) – n(A₁ÇA₄)

– n(A₂ÇA₃) – n(A₂ÇA₄) – n(A₃ÇA₄)+ n(A₁ Ç A₂ ÇA₃) + n(A₁ Ç A₂ ÇA₄)

+ n(A₁ Ç A₃ ÇA₄) + n(A₂ Ç A₃ ÇA₄) – n(A₁ÇA₂ Ç A₃ ÇA₄).

Example.1

Find the number of integers between 1 and 1000, both inclusive

(a) Which are divisible by either of 10, 15 and 25. .

(b) Which are divisible by neither 10 nor 15 nor 25.

Solution:

Let U denote the set of numbers between 1 and 1000 so that n(U) = 1000. Let A,B and C denote the sets of integers divisible by 10, 15 and 25 respectively.

Then n(A) = =100,

n(B) = = 66, and n(C) = = 40 .

S₁ = n(A) +n(B) +n(C) = 100 + 66 + 40 = 206

Considering two of the three sets A, B, C, we have

A Ç B = set of integers divisble by 10 and 15

= set of integers divisble by 30 (L.C.M of 10 and 15).

&⇒ n( A Ç B) = .

Similarly, B Ç C = set of integers divisble by 15 and 25

= set of integers divisble by 75 (L.C.M of 15 and 25).

&⇒ n( B Ç C) = .

Similarly, A Ç C = set of integers divisble by 10 and 25

= set of integers divisble by 50 (L.C.M of 10 and 25).

&⇒ n( A Ç C) = .

S₂ = 33 + 13 + 20 = 66.

Finally, A Ç B Ç C = set of integers divisble by 10, 15 and 25

= set of integers divisble by 150 (L.C.M of 10, 15 and 25).

&⇒ n( A Ç B Ç C) = .

S₃ = 6 .

(a) We have to find the number of integers divisible by either of 10, 15, and 25 i.e. n(AÈ BÈC).

n(AÈ BÈC) = S₁ – S₂ +S₃ = 206 – 66 + 6 = 146 .

(b) The number of integers divisible by neither 10 nor 15, nor 25

= n(U) – n(AÈ BÈC) = 1000 – 146 = 854.

Remark :

q The number of integers from 1 to n which are divisible by k is ([ .] denotes the greatest integer function). e.g. the number of integers from 1 to 100 which are divisible by 2 and 3 individually are and = 33 respectively.