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?
~~~~~~~~~~~~~



            The solution by other tutor is INCORRECT.


            I came to bring you the correct solution.


            Watch attentively every my step below.



<pre>
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.     <U>ANSWER</U>
</pre>

Solved   &nbsp;&nbsp;&nbsp;&nbsp;// &nbsp;&nbsp;&nbsp;&nbsp;in the RIGHT WAY.



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



It is how this problem &nbsp;SHOULD &nbsp;BE &nbsp;solved.


It is how this problem &nbsp;IS &nbsp;EXPECTED &nbsp;to be solved.



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


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



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


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