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
Given, length = 90 feet and Breadth = 66 feet
Stimulated length of tree plantation = 90-(5+5)=90-10 = 80 feet
Stimulated breadth of tree plantation = 66-(5+5)=66-10=56 feet
The number of columns bc = [leaving 4 feet in rows and columns]
=20+1=21
And the number of rows be = [leaving 4 feet in rows and columns]
== 14+1=15
Maximum number of trees that can be planted = 21 15 = 315