Question

n(AUB)=30 ; n(A-B)=12 ; n(B-A)=7 cuanto es n(A) + n(B)

Answer 1

Primero propiedades:
N(AuB)= n(A) + n(B) -n(A∩B) = 30
N (A-B)= n(A) - n(A∩B) = 12
(A∩B)= (B∩A)

30= n(A) - n (A∩B) +n (B)
30= 12+n (B)→ 18= n(B)

N (B-A)= n(b) - n(B∩A)
7= 18 -n (B∩A)
11= n(B∩A)  → B∩A=A∩B


30= n(A) - n (A∩B) +n (B)
30= n(A) - 11 + 18
30= n(A) +7
23= n(A)

n(A) + n(B)= 23+18=41