We will use the formulas # (1) : n(AuuB)=n(A)+n(B)-n(AnnB)#,
# (2) : n(A-AnnB) =n(A)-n(AnnB)#.
# (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.!
