Question

If A={1,2,3,4,5} B={4,5,6,7} then show that n(AUB)+nAnB) =n(A)+n(B)​

Answer 1

Step-by-step explanation:

Given :-

A = {1,2,3,4,5}

B = {4,5,6,7}

To find :-

Show that n (AuB) +n (AnB)=n(A)+n(B)

Solution :-

Given sets are :

A = {1,2,3,4,5}

The elements in A = 5

Therefore, n(A) = 5

B = {4,5,6,7}

The elements in B = 4

Therefore, n(B) = 4

and

AUB = {1,2,3,4,5} U {4,5,6,7}

AUB = { 1,2,3,4,5,6,7 }

The elements in AUB = 7

Therefore, n(AUB) = 7

and

AnB = {1,2,3,4,5} n {4,5,6,7}

AnB = { 4,5}

The elements in AnB = 2

Therefore, n(AnB) = 2

Now,

On taking LHS = n(AUB)+n(AnB)

=> 7+2

=> 9

n(AUB)+n(AnB) = 9 ---------------------(1)

On taking RHS = n(A)+n(B)

=> 5+4

=> 9

n(A)+n(B) = 9 ------------------------------(2)

From (1)&(2)

LHS = RHS

Therefore, n(AUB)+n(AnB) = n(A)+n(B)

Hence, Proved.

Answer :-

n(AUB)+n(AnB) = n(A)+n(B)

Used formulae:

The elements in the set A is denoted by n(A)

The fundamental theorem on sets

n(AUB) = n(A)+n(B) -n(AnB)

Answer 2

Answer:

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Step-by-step explanation:

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