Explanation:
Solution :-
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)
Verified .
Used formulae:-
→ 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.