SOLUTION: In a group of 120 students numbered 1 to 120, all even numbered students choose Physics, students whose numbers are divisible by 5 choose Chemistry and those whose numbers are divi

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Question 1176415: In a group of 120 students numbered 1 to 120, all even numbered students choose Physics, students whose numbers are divisible by 5 choose Chemistry and those whose numbers are divisible by 7 choose Economics. How many students choose none of the three subjects?
Found 2 solutions by ankor@dixie-net.com, ikleyn:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
In a group of 120 students numbered 1 to 120,
all even numbered students choose Physics,
120/2 = 60 students
students whose numbers are divisible by 5 choose Chemistry and
120/5 = 24 students
those whose numbers are divisible by 7 choose Economics.
120/7 ~ 17 students
How many students choose none of the three subjects?
120 - 60 - 24 - 17 = 19 students choose none of these

Answer by ikleyn(52835) About Me  (Show Source):
You can put this solution on YOUR website!
.
In a group of 120 students numbered 1 to 120,
all even numbered students choose Physics,
students whose numbers are divisible by 5 choose Chemistry
and those whose numbers are divisible by 7 choose Economics.
How many students choose none of the three subjects?
~~~~~~~~~~~~~


            The solution by other tutor is INCORRECT.

            I came to bring you the correct solution.

            Watch attentively every my step below.


We have the sets

    P of  120/2 = 60 students   (Physics)

    C of  120/5 = 24 students   (Chemistry)

    E of  [120/7] = 17 students (Economics)



We have their in-pair intersections

    PC of  120/(2*5)   = 12 students
 
    PE of  [120/(2*7)] =  8 students
 
    CE of  [120/(5*7)] =  3 students.


We have their triple intersection

    PCE of 120/(2*5*7) =  1 student.



Now we apply the inclusive-exclusive formula to find the number of students in the UNION of sets (P U C U E)


    n(P U C U E) = n(P) + n(C) + n(E) - n(PC) - n(PE) - n(CE) + n(PCE)  (the alternate sum)

                 = 60 + 24 + 17 - 12 - 8 - 3 + 1 = 79.


The rest of the students,  120 - 79 = 41,  choose NONE of the three subjects.     ANSWER

Solved     //     in the RIGHT WAY.


/\/\/\/\/\/\/\/


It is how this problem  SHOULD  BE  solved.

It is how this problem  IS  EXPECTED  to be solved.


The solution of the other tutor is a  PERFECT  EXAMPLE  of how this problem  SHOULD  NOT  be solved.

It is classic example of the typical  ERROR  which  EVERYBODY  MAKES  who is unfamiliar with the right approach.


Now you should be  ABSOLUTELY  HAPPY,  because after my post you  KNOW  BOTH

how this problem  SHOULD  BE  solved and how you  SHOULD  NOT  even try to approach it.