From the following Venn diagram, write the elements of the set A and B and verify n(AUB) + n(A∩B) = n(A) + n(B)

n(A)+n(B)

Step-by-step explanation:

Step-by-step explanation:

From the given Venn diagram ,

A = { a,c,d,f,n }

Number of elements in the set A = 5

=> n(A) = 5

B = { a,b,d,e,g,n }

Number of elements in the set B = 56

=> n(B) = 6

AUB = { a,b,c,d,e,f,g,n}

Number of elements in the set AUB = 8

=> n(AUB) = 8

AUB = { a,d,n}

Number of elements in the set AnB = 3

=> n(AnB) = 3

Now

n(AUB)+n(AnB)

=> 8+3

=> 11

n(AUB)+n(AnB) = 11 (1)

On taking n(A)+n(B)

=> 5+6

=> 11

n(A)+n(B) = 11 (2)

From (1)&(2)

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

→ Fundamental Theorem on sets

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

→ The set of all elements in either A or in B or in both A and B is AUB.

→ The set of all common elements in both A and B is AnB.

→ The number of elements in A is denoted by n(A).

Hope this helps.

5

step-by-step explanation:

c.

step-by-step explanation: