In a class of $40$ students, $12$ enrolled for both English and German. $22$ enrolled for German. If the students of the class enrolled for at least one of the two subjects, then the number of students who enrolled for English but not German is

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Solution

Let $E$ be the set of students enrolled in English
and $G$ be the set of students enrolled in German
$\Rightarrow n (E \cup G) = 40, n (E) = 30, n (E \cap G) = 12$
Now, set of students who enrolled only in english or $N - G$
We have $n (E - G) = n (E) - n (E \cap G) = 30 - 12 = 18$

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In a class of 40 students, 12 enrolled for both English and German. 22 enrolled for German. If the students of the class enrolled for at least one of the two subjects, then the number of students who enrolled for English but not German is

Let E be the set of students enrolled in English and G be the set of students enrolled in German ⇒n(E∪G)=40,n(E)=30,n(E∩G)=12 Now, set of students who enrolled only in english or N−G We have n(E–G)=n(E)–n(E∩G)=30−12=18

In a class of $40$ students, $12$ enrolled for both English and German. $22$ enrolled for German. If the students of the class enrolled for at least one of the two subjects, then the number of students who enrolled for English but not German is

Let $E$ be the set of students enrolled in English
and $G$ be the set of students enrolled in German
$\Rightarrow n (E \cup G) = 40, n (E) = 30, n (E \cap G) = 12$
Now, set of students who enrolled only in english or $N - G$
We have $n (E - G) = n (E) - n (E \cap G) = 30 - 12 = 18$