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This question in my text book chapter named "Permutation, Combination and Probability". But I am stuck with Permutation, Combination and probability. All things are seems same to me. As I am new in these.

Question is: There are 53 students taking Chemistry and Physics or both, 38 students taking chemistry, and 40 students taking physics. (i) How many students are taking both Chemistry and Physics? (ii) How many students are taking Chemistry but not Physics? (iii) How many students are taking Physics but not Chemistry?

Can any one help me out.

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  • $\begingroup$ "Chemistry and Physics or both" doesn't make any sense. Perhaps the question was supposed to read, "53 students taking Chemistry OR Physics or both". $\endgroup$ – Gerry Myerson Jun 23 '14 at 7:12
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    $\begingroup$ (i) $38+40-53=25$. (ii) $38-25=13$. (iii) $40-25=15$. $\endgroup$ – barak manos Jun 23 '14 at 7:28
  • $\begingroup$ Yes the correct answers. $\endgroup$ – user159627 Jun 23 '14 at 7:29
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This is an elementary set theoretic problem. Let $P$ be the set of students taking Physics and $C$ be the set of students taking Chemistry. Then $P\cup C$ is the set of students taking either Physics or Chemistry or both. And $P\cap C$ is the set of students taking both Physics and chemistry. Also $P\setminus C$ is the set of students taking Physics but not Chemistry. Now, we denote the number of elements in a set $A$ by $n(A)$. The following relation will be useful for your answer:

$$n(P\cup C)=n(P)+n(C)-n(P\cap C)$$

$$n(P)=n(P\setminus C)+n(P\cap C)$$

So, in your problem, $n(P)=40, n(C)=38, n(P\cup C)=53$. So you can calculate the rest.

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    $\begingroup$ Thank you so much. I am going to solve it using above relation. $\endgroup$ – user159627 Jun 23 '14 at 7:26
  • $\begingroup$ Yes done with it. $\endgroup$ – user159627 Jun 23 '14 at 8:02
  • $\begingroup$ But i did not get the notification .. did you click the Tick mark on the left side of the answer ? $\endgroup$ – Debashish Jun 23 '14 at 8:04
  • $\begingroup$ @user159627 .. thanks !! $\endgroup$ – Debashish Jun 23 '14 at 8:19
  • $\begingroup$ done Debashish. $\endgroup$ – user159627 Jun 23 '14 at 8:21

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