Let A=favoured proposal III
B=favoured proposal III
C=favoured proposal III
Here, n(A U B U C)=78
We Knows,
n(AUBUC)= n(A)+n(B)+n(C)-n(A ∩ B)-n(A ∩ C)-n(A ∩ B)+n(A ∩ B ∩ C)
78 = 50 + 30 + 20 - n(A ∩ B) - n(A ∩ C) - n(A ∩ B) + 5.
n(A ∩ B) + n(A ∩ C) + n(A ∩ B) = 27
For those who favored exactly 2 proposals we have to subtract n(A ∩ B ∩ C) from n(A ∩ B) , n(A ∩ C) and n(A ∩ B)
n(A ∩ B) + n(A ∩ C) + n(A ∩ B) - 3*n(A ∩ B ∩ C)
=27-15 =12
For those who favored exactly 3 proposals=5
Now, those who favored more than one of the 3 proposals = 12 + 5 =17
Hence,Option (C)17 is the correct choice.