Question

From the given Venn diagram , verify n(AUB) + n(AUB) = n( A) + n(B).​

Answer 1

Correct Question:-

Given to verify :-

n(AUB) + n(A∩B) = n( A) + n(B)

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➣ Explanation:-

Take L.H.S

n(AUB) + n(AUB)

According to diagram ,

(AUB) it is read is A union B it means We have to take all the elements in the set . So,

(AUB) = { 2, 4, 3, 1, 6,5}

(AUB) = {1, 2, 3, 4, 5,6 }

n(AUB) means number of elements present in (AUB)

So,

n(AUB) = 6

(A∩B) It can be read as A intersection B it means which is common element in set .Here the common element is

(A∩B) = {3}

n(A∩B) means number of elements present in (A∩B )

n(A∩B)= 1

n(AUB) + n(A∩B) = 6 + 1

n(AUB) + n(A∩B) = 7

L.H.S = 7

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Now take R.H.S ,

n( A) + n(B)

Firstly lets find (A) , (B)

Here the common element is 3 i.e it present in both sets So,

Elements in A is = {2,3,4}

Elements in B is = {1, 6,5,3}

So, total number of elements in both sets i.e n(A) , n(B) is

n(A) = 3

n(B) = 4

n(A) + n(B) = 3+4

n(A) + n(B) =7

So, R.H.S = 7

L.H.S = R.H.S = 7

So,

n(AUB) + n(A∩B) = n( A) + n(B)

Hence verified ! ${}$

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Answer 2

Answer:

Correct Question:-

Given to verify :-

$\huge \: n(AUB) + n(A∩B) = n( A) + n(B)$

➣ Explanation:-

Take L.H.S

$\: n(AUB) + n(AUB)$

According to diagram ,

(AUB) it is read is A union B it means We have to take all the elements in the set . So,

$= > (AUB) = { 2, 4, 3, 1, 6,5}$

$= > \: (AUB) = {1, 2, 3, 4, 5,6 }$

n(AUB) means number of elements present in (AUB)

So,

$\sf \: {n(AUB) = 6}$

(A∩B) It can be read as A intersection B it means which is common element in set .Here the common element is

$\sf \: {(A∩B) = {3}}$

n(A∩B) means number of elements present in (A∩B )

n(A∩B)= 1

n(AUB) + n(A∩B) = 6 + 1

n(AUB) + n(A∩B) = 7

L.H.S = 7

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Now take R.H.S ,

n( A) + n(B)

Firstly lets find (A) , (B)

Here the common element is 3 i.e it present in both sets So,

Elements in A is = {2,3,4}

Elements in B is = {1, 6,5,3}

So, total number of elements in both sets i.e n(A) , n(B) is

n(A) = 3

n(B) = 4

n(A) + n(B) = 3+4

n(A) + n(B) =7

So, R.H.S = 7

L.H.S = R.H.S = 7

So,

n(AUB) + n(A∩B) = n( A) + n(B)