Here we are going to discuss rules for the elements of a finite set.

**At first, we have to know what is this.**## Rules for the elements of a finite set

Let a set is P for

Here the number of elements of set P is = 6.

Elements of set P are represented with n(P). So, in this case n(P) = 6.

Now again, let A = {1, 2, 3, 4} and B = {3, 4, 5, 6}

So, n(A) + n(B) = 4 +4 = 8

Also, union of set A and set B is A∪B = {1, 2, 3, 4, 5, 6}

So, n(A∪B) = 6

Now intersection of set A and set B is A∩B = {3, 4}

So, n(A∩B) = 2

Carefully see that in n(A) + n(B) common elements are counted two times but we know those common elements of set A and set B are represented by A∩B. So, from this, we can write,

Total number of elements of first set + second set = Union of two sets + Intersection of those two sets

n(A) + n(B) = n(A∪B) + n(A∩B)

∴

If there are no common elements between set A and set B then

n(A∪B∪C)

= n{ (A∪B)∪C}[From the set-associative law we get, A∪B∪C = (A∪B)∪C]

= n(A∪B)+n(C) - n{(A∪B)∩C}

= n(A) + n(B) − n(A∩B) + n(C) - n{(A∩C)∪(B∩C)}

= n(A) + n(B) − n(A∩B) + n(C) - {n(A∩C)+n(B∩C)-n(A∩B∩C)}

= n(A) + n(B)+ n(C)+n(A∩B∩C)- {n(A∩B)+n(B∩C)+n(C∩A)}

∴n(A∪B∪C) = {n(A) + n(B) +n(C)+n(A∩B∩C)}-{n(A∩B)+n(B∩C)+n(C∩A)}

**Rules for the elements of a finite set**and P = {1, 2, 3, 4, 5, 6}Here the number of elements of set P is = 6.

Elements of set P are represented with n(P). So, in this case n(P) = 6.

Now again, let A = {1, 2, 3, 4} and B = {3, 4, 5, 6}

So, n(A) + n(B) = 4 +4 = 8

Also, union of set A and set B is A∪B = {1, 2, 3, 4, 5, 6}

So, n(A∪B) = 6

Now intersection of set A and set B is A∩B = {3, 4}

So, n(A∩B) = 2

### Rule 01: n(A∪B) = n(A) + n(B) − n(A∩B)

Carefully see that in n(A) + n(B) common elements are counted two times but we know those common elements of set A and set B are represented by A∩B. So, from this, we can write,

Total number of elements of first set + second set = Union of two sets + Intersection of those two sets

n(A) + n(B) = n(A∪B) + n(A∩B)

∴

**n(A∪B) = n(A) + n(B) − n(A∩B)**If there are no common elements between set A and set B then

**n(A∪B) = n(A) + n(B) − 0 =****n(A) + n(B)**### Rule 02: n(A∪B∪C) = {n(A) + n(B) +n(C)+n(A∩B∩C)}-{n(A∩B)+n(B∩C)+n(C∩A)}

n(A∪B∪C)

= n{ (A∪B)∪C}[From the set-associative law we get, A∪B∪C = (A∪B)∪C]

= n(A∪B)+n(C) - n{(A∪B)∩C}

**[From the rule 1 n(A∪B) = n(A) + n(B) − n(A∩B)]**= n(A) + n(B) − n(A∩B) + n(C) - n{(A∩C)∪(B∩C)}

**[From the rule 1 n(A∪B) = n(A) + n(B) − n(A∩B)**

**and**

**From the distributive law we get (A∪B)∩C=(A∩C)∪(B∩C)]**

= n(A) + n(B) − n(A∩B) + n(C) - {n(A∩C)+n(B∩C)-n(A∩B∩C)}

**[Again using rule 1]**= n(A) + n(B)+ n(C)+n(A∩B∩C)- {n(A∩B)+n(B∩C)+n(C∩A)}

∴n(A∪B∪C) = {n(A) + n(B) +n(C)+n(A∩B∩C)}-{n(A∩B)+n(B∩C)+n(C∩A)}

## 0 Comments