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

$n (E \cup G) = 40 \Rightarrow n (E) = 30 ∴ n (E - G) = n (E) - n (E \cap G) = 30 - 12 = 18$

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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

n(E∪G)=40⇒n(E)=30∴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

$n (E \cup G) = 40 \Rightarrow n (E) = 30 ∴ n (E - G) = n (E) - n (E \cap G) = 30 - 12 = 18$