Question 1149313
.
In a class of 36 students 25 study chemistry and 22 study mathematics and 25 study physics ,17 study physics and mathematics,
18 study physics and chemistry,15 study one of the three subject 
Find a)Those who study all three subject b) only mathematics and chemistry 
~~~~~~~~~~~~~~~~


            The solution by @greenestamps is correct;         (G)

            The solution by @math_helper is correct;            (MH)

            The solution by Edwin is  NOT  CORRECT.               (E)


            My solution is  BELOW.   It supports  (G)  and  (MH)  and disproves  (E).


            See also my comments after my solution.



<pre>
We are given a universal set of 36 students (the entire class).


We are also given info about three its subsets : 

    C (students enrolled to Chemistry)  of n(C) = 25 elements;  
    M (Mathematics)                     of n(M) = 22 elements, and 
    P (Physics)                         of n(P) = 25 elements.


We are also given the following info about their in-pair intersections:

    PM (studying Physics and Math)      of n(PM) = 17 elements;
    PC (studying Physics and Chemistry) of n(PC) = 18 elements;
    MC (studying Math    and Chemistry) of n(MC) = unknown number of elements.

 
We also should find the number of elements in the the triple intersection

    PMC (studying Physics, Math and Chem) of n(PMC) (unknown number of elements).


We assume (from the <U>context</U>) that the union of subsets P, M and C  is the entire set of the 36 students in the class.


In this situation the following formula works


    n(P U M U C) = n(P) + n(M) + n(C) - n(P n M) - n(P n C) - n(M n C) + n(P n M n C).     (1)


Substitute the given data. You will get


    36 = 25 + 22 + 25 - 17 - 18 - n(MC) + n(PMC),   or

    n(MC) - n(PMC) = 25 + 22 + 25 - 17 - 18 - 36 = 1.


This difference, n(MC) - n(PMC), is exactly the number of those who study only Math and Chemistry,

so it is just answer to question b) of the problem.


Now we will work to answer question a) and to find the unknown n(PMC).

For it, I should create the other equation, based on the second given condition.


The number of those who study only Math      is  22-n(PM)-n(MC)+n(PMC) = 22-n(PM)-(n(MC)-n(PMC)) =
                        replace here n(PM) by 17 and replace (n(MC)-n(PMC)) by 1, as we just found, to continue the line
                                                                                                 = 22-17-1 = 4.

The number of those who study only Physics   is  25-n(PM)-n(PC)+n(PMC) = 25-17-18+n(PMC) = -10+n(PMC).

The number of those who study only Chemistry is  25-n(PC)-n(MC)+n(PMC) = 25-n(PC)-(nMC)-n(PMC) = 
                        replace here n(PC) by 18 and replace (n(MC)-n(PMC)) by 1, as we just found, to continue the line                   
                                                                                               = 25-18-1 = 6.

Now the equation "the sum of those who study only one of the three subjects" is

    4 - 10+n(PMC) + 6 = 15,  which implies  n(PMC) = 15 - 4 + 10 - 6 = 15.


<U>ANSWER to (a)</U> : 15.    <U>ANSWER to (b)</U> : 1.
</pre>

Solved.


-----------------


The last step is to prove the formula (1).


<pre>
    It is totally clear to you why I add the first three addends in the formula (1).

    But when I add them, I count twice every term in each in-pair intersection.

    Therefore, I subtract the numbers of terms in each in-pair intersection.

    Next, when I add three first addends, I count thrice each term in the triple intersection;

    and when I subtract in-pair intersections, I cancel these terms thrice.

    Therefore, I must add the number of terms in the triple intersection one more time to restore the balance.
</pre>

Thus the formula &nbsp;(1) &nbsp;is proved &nbsp;&nbsp;// &nbsp;&nbsp;and the solution is completed &nbsp;(!&nbsp;!)



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


My comments.


<pre>
    Probably, the formulation is not perfect; probably, it could be better.

    But as formulated, it does not leave any doubt/doubts

        - that the condition covers and involves all 36 students of the class, and
        - what does it mean "15 study one of the three subject".


     It is a STANDARD WAY to formulate such circumstances in this sort of problems,
     and I saw tens of them in this forum, formulated in this way.



     Regarding matrix solutions by Edwin and by @Math_helper, I do not think that it is the expected way for the school students
     to get the solution.


     It is why I produced my own solution and placed it here.
</pre>


Thank you for your attention (!)



===================


Dear reader, if you want to see other similar solved problems, to wider your horizon and to feel 
the solid ground under your legs, look into the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/misc/Counting-elements-in-sub-sets-of-a-given-finite-set.lesson>Counting elements in sub-sets of a given finite set</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Advanced-probs-counting-elements-in-sub-sets-of-a-given-finite-set.lesson>Advanced problems on counting elements in sub-sets of a given finite set</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Challenging-problems-on-counting-elements-in-subsets-of-a-given-finite-set.lesson>Challenging problems on counting elements in subsets of a given finite set</A> 

in this site.


Look also into the links


https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1132870.html

https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1129554.html

https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1129554.html

https://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Rate-of-work-word-problems.faq.question.1126097.html

https://www.algebra.com/algebra/homework/word/evaluation/Evaluation_Word_Problems.faq.question.1126099.html


to similar solved problems in the archive of this forum.



When you learn from these sources, you will know the subject as well, as I know it . . .