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.!