You are given the universal set U of 70 students and 3 its basic subsets P, B and C 
(see the table below).

    U    70    total students

    P    23    Physics

    B    25    Biology

    C    22    Chemistry       


Also, you are given info about their in-pair intersections and about their triple intersection.


    PB    6    Physics and Biology

    BC    7    Biology and Chemistry  

    PC    8    Chemistry and Physics

    PBC   2    Physics, Biology and chemistry



Having this info well organized, you can easily answer all questions (a), and (b). 



(a)  The set of students not taking any of the three subjects is  U \ (P U B U C).
     
     So, calculate the number of students in the union (P U B U C}  first.

     For it, use the inclusion-exclusion princuple/(formula)

         n(P U B U C) = n(P) + n(B) + n(C) - n(PB) - n(BC) - n(PC) + n(PBC) = 

                      =  23  +  25  +  22  -  6   -   7   -   8   +  2    = 51.


     Now the next step gives the answer to question (a) :  

        the number of students taking none of the three sciences = 70 - 51 = 19.    ANSWER



(b)  The number of students taking Physics only, n(Po), is  

         n(Po) = n(P) - n(PB) - n(PC) + n(PBC) = 23 - 6 - 8 + 2 = 11.


     The number of students taking Biology only, n(Bo), is  

         n(Bo) = n(B) - n(PB) - n(BC) + n(PBC) = 25 - 6 - 7 + 2 = 14.


     The number of students taking Chemistry only, n(Co), is  

         n(Co) = n(C) - n(PC) - n(BC) + n(PBC) = 22 - 8 - 7 + 2 = 9.


     The number of students taking just one of the three subjects is

         n(Po) + n(Bo) + n(Co) = 11 + 14 + 9 = 34.