Answer:
Part (a) n(A∩B∩C) = 4
Part (b) n(A∪B∪C) = 54
Explanation:
n(A) = no. of students who took Math 2A
n(B) = no. of students who took Math 2B
n(C) = no. of students who took Math 2C
n(A∩B) = no. of students who took both Math 2A and 2B
n(A∩C) = no. of students who took both Math 2A and 2C
n(B∩C) = no. of students who took both Math 2B and 2C
n(A∩B∩C) = no. of students who took all three Math 2A, 2B and 2C
n(A∪B∪C) = no. of total students in a group
∩ represents Intersection and ∪ represents Union
Part (a)
n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(A∩C) - n(B∩C) + n(A∩B∩C)
Where n(A∩B∩C) represents the number of students who took all three classes and n(A∪B∪C) represents the total number of students in group A
157 = 51 + 80 + 70 - 15 - 20 - 13 + n(A∩B∩C)
Re-arranging the equation to solve for n(A∩B∩C) since we want to find out those students who took all three classes
n(A∩B∩C) = 157 - 51 - 80 - 70 + 15 + 20 + 13
n(A∩B∩C) = 4
So there are 4 students in group A who took all three classes
Part (b)
n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(A∩C) - n(B∩C) + n(A∩B∩C)
This time we are given n(A∩B∩C) students who took all three classes and want to find n(A∪B∪C) that is total number of students
n(A∪B∪C) = 28 + 28 + 25 - 11 - 9 - 10 + 3
n(A∪B∪C) = 54
So there are total 54 students in group B
