Since n(C∩A)≥n(A∩B∩C) We have n(C∩A)−n(A∩B∩C)≥0...(ii) From (i) and (ii) n(A∪B∪C)≤28....(iii) Now,
n(A∪B)=n(A)+n(B)−n(A∩B)
=10+15−8=17
and=n(B∪C)=n(B)+n(C)−n(B∩C)
=15+20−9=26 Since n(A∪B∪C)≥n(A∪C) and n(A∪B∪C)≥n(B∪C), we have n(A∪B∪C)≥17 and n(A∪B∪C)≥26 Hence n(A∪B∪C)≥26...(iv) From (iii) and (iv) we obtain 26≤n(A∪B∪C)≤28 Also n(A∪B∪C) is a positive integer ∴n(A∪B∪C)=26or27or28