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## Here's the Solution to this Question

Let the $J, C, P$ be the sets of those that took Java, Calculus and Precalculus courses, respectively.

It is know that the overall amount of students is 50.

the amount of students in $P = 30$

the amount of students in $J\bigcap P = 16$

the amount of students in $C = 18$

the amount of students in $J\bigcap C = 8$

the amount of students in $J = 26$

the amount of students in $J\bigcup C \bigcup P = 47$

the amount of students in $C\bigcap P = 9$

A. Since the are 50 students and 47 took at least one course, there are $50 -47 = 3$ such students

B. $J\bigcup C \bigcup P = J + C + P +$

$+(- J\bigcap C - J\bigcap P -C\bigcap P + J\bigcap C \bigcap P) =$

$= 26 + 18 + 30 - 8 - 16 - 9 + J\bigcap C \bigcap P =$

$= 41 + J\bigcap C \bigcap P$

$47 = 41 + J\bigcap C \bigcap P$

$J\bigcap C \bigcap P = 6$

6 students took both Java, Calculus and Precalculus courses.

C. $(C\bigcap P)\setminus J = C\bigcap P - C\bigcap P \bigcap J = 9 - 6 = 3$

3 students took Precalculus and Calculus but not Java.

$P\setminus (C\bigcup J) = P - P\bigcap(C\bigcup J)=$

$=P - (P\bigcap C)\bigcup (P\bigcap J).$

$(P\bigcap C)\bigcup (P\bigcap J) =$

$= P\bigcap C + P\bigcap J -(P\bigcap C)\bigcap (P\bigcap J) =$

$= P\bigcap C + P\bigcap J -P\bigcap C\bigcap J =$

$=$ 9 + 16 - 6 = 19

$P - (P\bigcap C)\bigcup (P\bigcap J) = 30 - 19 = 11 =$

$=P\setminus (C\bigcup J)$

11 students took precalculus but neither calculus nor Java