We will use the formulas (1) : n(AuuB)=n(A)+n(B)-n(AnnB)(1):n(A∪B)=n(A)+n(B)−n(A∩B),
(2) : n(A-AnnB) =n(A)-n(AnnB)(2):n(A−A∩B)=n(A)−n(A∩B).
(3) : (AuuB)'=A'nnB'...............[De'Morgan's law].
where, A & B sub U and, n(A) denotes the Number of Elements in a Set A sub U, the Universal Set .
Let M= The Set of students taking Maths. , and, S that of
students taking Sc.. Hence, MnnS is the set of students taking both the subjects, whereas, M'nnS' is the set of students opting neither of the subjects.
Our goal is to find n(M'nnS')=n((MuuS)'), because of (3).
Now, let us observe that, (M-MnnS)uu(MnnS)=M and, their intersection is phi, so, by (1), we get,
n(M)=n(M-MnnS)+n(MnnS), i.e.,
242=n(M-MnnS)+183 rArr n(M-MnnS)=59.
(2) rArr n(M)-n(MnnS)=59,
From (1), then, we have,
n(MuuS)=n(M)+n(S)-n(MnnS)=59+208=267
Therefore, n(M'nnS')=n((MuuS)')=n(U-MuuS)
=n(U)-n(MuuS)=300-267=33.
Enjoy Maths.!