Answer: The answer is 4 and 32.
Step-by-step explanation: Let "A", "B" and "C" represents the set of students who were taking Arabic, Bulgarian and Chinese respectively.
The, according to the given information, we have
[tex]n(A)=33,~~n(B)=32,~~n(C)=40,~~n(A\cap B)=14.[/tex]
Let 'p' represents the number of students who take all the three languages, then
[tex]n(A\cap B\cap C)=p.[/tex]
Also,
[tex]n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)=9\\\\\Rightarrow n(A\cap B)+n(A\cap C)=24+p~~~~~~~~~~~~~~(a),\\\\n(A\cap B)+n(B\cap C)=20+p~~~~~~~~~~~~~~(b),\\\\n(B\cap C)+n(A\cap C)=20+p~~~~~~~~~~~~~~~(c).[/tex]
From here, we get after subtracting equation(c) from (b) that
[tex]n(A\cap B)=n(A\cap C)=14.[/tex]
Therefore,
[tex]p=14+14-24=4,[/tex] and from equation (a), we find
[tex]n(B\cap C)=24-14=10.[/tex]
Thus,
[tex]n(A\cap B\cap C)=4[/tex] and
[tex]n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\\\\\Rightarrow n(A\cap B\cap C)=33+32+40-9-12-20+4\\\\\Rightarrow n(A\cap B\cap C)=68.[/tex]
Thus, the number of students who take all the three languages is 4 and the number of students who take none of the languages is 100-68 = 32.
