Given n(U)=200,n(M)=120(p)=90, n(C)=70;n(M∩P)=40,n(P∩C)=30 n(C∩M)=50,n(M′∩P′∩C′)=20
Since, n(M′∩P′∩C′)=20
⇒n(M∪P∪C)′=20 (by Demorgan's law)
⇒n(U)−n(M∪P∪C)=20
⇒n(M∪P∪C)=180.
⇒n(M)+n(P)+n(C)−n(M∩P)−n(P∩C)−n(C∩M)+n(M∩p∩C)
∴180=(120+90+70)−(40+30+50)+n(M∩P∩C)
⇒n(M∩P∩C)=20.
Hence 20 students studied all the three subjects.