In a certain college of the 100 students, 60% students took Finance course, 45% took Marketing course and 45 students took Operation Management course. If 10 students took all the courses and 10 students none of the three courses, how many students took exactly 2 of these 3 courses?
(Set)Here, to solve the problem, we need to use some laws from the primary concept of sets.
Let, n(F) = 60; n(M) = 45 n(Om)=45; n(T) = 100
All courses taken, n(F M Om )=10
None taken, n(F M Om )^ =10
We know, z(F M Om )=n(T)-n(F M Om)^
= n(F M Om)= 100 - 10 = 90
= So, we can write, n (F M Om)
= n(F)+n(M)+n(Om)-n(FM)-n(MOm)-n(OmF)+n(FMOm)
= 90=60+45+45-{n(FM)+n(MOm)+n(OmF}+10
=90=160-ln(F M)+n(M Om )+n(Om F)
n(F M)+n(M Om )+n(Om F)=70
The number of students who took 2 of these 3 courses is
=70-3(F M Om)= 70 - (3 10) = 70 - 30 = 40