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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?
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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.
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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.