Given,
Let P denotes Physics
M denotes Mathematics
and C denotes Chemistry
n(P)=20n(M)=24n(C)=22n(P∩M)=10n(P∩C)=12n(M∪P∪C)=40
Now, 16 had only 2 of the three books
that means
n(P∩M)+n(P∩C)+n(M∩C)−2n(M∩P∩C)=16⇒10+12+n(M∩C)2n(M∩P∩C)=16⇒2n(M∩P∩C)−n(M∩C)=6 ......(1)
and we know that,
n(M∪P∪C)=n(P)+n(M)+n(C)−n(P∩C)−n(P∩M)−n(M∩C)+n(M∩P∩C) 40=20+24+22−12−10−n(M∩C)+n(M∩P∩C)n(M∩P∩C)−n(M∩C)=−4 .......(2)
Solving equation (1) and (2)
we get
n(M∩P∩C)=10n(M∩C)=14
(a)Students that have all 3 books =n(P∩M∩C)=10
(b)Students that have only mathematics and chemistry books=n(M∩C)=14
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