A) 16
B) 6
C) 8
D) 20
Correct Answer: A
Solution :
Given\[n(N)=12\], \[n(P)=16\], \[n(H)=18\], \[n(N\cup P\cup H)=30\] From, \[n(N\cup P\cup H)=n(N)+n(P)+n(H)-n(N\cap P)\] \[-n(P\cap H)-n(N\cap H)+n(N\cap P\cap H)\] \ \[n(N\cap P)+n(P\cap H)+n(N\cap H)=16\] Now, number of pupils taking two subjects \[=n(N\cap P)+n(P\cap H)+n(N\cap H)-3n(N\cap P\cap H)\] \[=16-0=16\].You need to login to perform this action.
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