A class of 40 students were each to required three text book of Physics, Mathematics and chemistry however 20 students had physics books, 24 had Mathematics books and 22 had chemistry books. 10 students had Physics and mathematics books ,12 students had physics and chemistry books and 16 had only 2 of the three books. if Every students had at least one of the three books, find, a. how many students had all 3 books. b. how many students only mathematics and chemistry books

Given,

Let P denotes Physics

M denotes Mathematics

and C denotes Chemistry

$n(P)=20\\n(M)=24\\n(C)=22\\n(P\cap M)=10\\n(P\cap C)=12\\n(M\cup P\cup C)=40$

Now, 16 had only 2 of the three books

that means

$n(P\cap M)+n(P\cap C)+n(M\cap C)-2n(M\cap P\cap C)=16\\\Rightarrow 10+12+n(M\cap C)2n(M\cap P\cap C)=16\\\Rightarrow 2n(M\cap P\cap C)-n(M\cap C)=6\ \ \ ......(1)$

and we know that,

$n(M\cup P\cup C)=n(P)+n(M)+n(C)-n(P\cap C)-n(P\cap M)-n(M\cap C)+n(M\cap P\cap C)\\\ \\40=20+24+22-12-10-n(M\cap C)+n(M\cap P \cap C)\\n(M\cap P\cap C)-n(M\cap C)=-4\ \ \ .......(2)$

Solving equation (1) and (2)

we get

$n(M\cap P\cap C)=10\\n(M\cap C)=14$

$(a)\text{Students that have all 3 books }= n(P\cap M\cap C)= 10$

$(b)\text{Students that have only mathematics and chemistry books}=n(M\cap C)=14$