# How many students studied chemistry, given that they all also took physics?

Q:A class of 25 students has options as to which science subjects they study. The options are biology, chemistry, geology and physics. 15 of the 25 students took only two of these sciences, whilst the other 10 students took three. 5 students studied geology and 8 studied biology, but nobody took both. How many students studied chemistry, given that they all also took physics?

I know that n(G)=5,n(B)=8,n(BnG)=0,n(BnCnGnP)=0, there is no one who takes only 1 subject and no one who takes all 4 subjects. Students who take chemistry will also take physics.I also know that n(CnP)=n(BnCnP)+n(PnCnG)=10. So how can I get the n(C)?

Let's state Biology as B, Chemistry as C, Geology as G and Physics as P.

What is given -

$$N(T) = 25, N(B) = 8, N(G) = 5, N(P) = 25, N(B \cap G) = 0$$, $$T$$ is total students.

We also know $$10$$ students take $$3$$ of these $$4$$ subjects and $$15$$ take $$2$$ subjects.

So all $$12 \, ( = 25-8-5) \,$$ students who take neither G nor B must have taken both P and C.

$$N(P \cap C \cap \neg(B \cup G)) = 12$$

As $$10$$ students take $$3$$ subjects and nobody takes both B and G, those $$10$$ students must be taking P, C and either B or G.

$$N(P \cap C \cap (B \cup G)) = 10$$

Other $$3$$ out of $$13$$ taking B or G must not be taking C as they can take only two and they all take Physics.

$$N(P \cap \neg C \cap (B \cup G)) = 3$$

So you know number of students taking Chemistry is $$22$$.