Answer to Question #176481 in Discrete Mathematics for Isaac Aidoo

Question

In a class of 35 students it is known that 24 of them do arts, 20 do chemistry and 22 do biology. All the students do at least one of the 3 subjects, 3 do all the three subjects while 7 do art and biology. 6 do art and chemistry but not biology and 8 do chemistry and biology. How many of them do chemistry or biology only or arts only?

Answer 1

In a class of 45 students 

$$N(A\cup C\cup B)=45$$

It is known that 24 of them do arts, 20 do chemistry and 22 do biology. 

$$N(A)=24, N(C)=20, N(B)=22$$

3 do all the three subjects

$$N(A\cap C\cap B)=3$$

7 do art and biology

$$N(A\cap B)=7$$

6 do art and chemistry but not biology

$$N(A\cap C \cap B^C)=6$$

8 do chemistry and biology

$$N(C\cap B)=8$$

Then

$$N(A\cap B \cap C^C)=N(A\cap B)-N(A\cap B\cap C)$$

$$=7-3=4$$

$$N(B\cap C\cap A^C)=N(B\cap C)-N(A\cap B\cap C)$$

$$8-3=5$$

$$N(A\cap B^C\cap C^C)=N(A)-N(A\cap B\cap C^C)$$

$$-N(A\cap C\cap B^C)-N(A\cap B\cap C)$$

$$=24-4-6-3=11$$

$$N(B\cap A^C\cap C^C)=N(B)-N(B\cap A\cap C^C)$$

$$-N(B\cap C\cap A^C)-N(A\cap B\cap C)$$

$$=22-4-5-3=10$$

$$N(C\cap A^C\cap B^C)=N(C)-N(C\cap A\cap B^C)$$

$$-N(C\cap B\cap A^C)-N(A\cap B\cap C)$$

$$=20-6-5-3=6$$

Check

$$11+10+6+4+6+5+3=45$$

$N(\text{chemistry or biology only})=N(C\cap B\cap A^C)=5$

$N(\text{arts only })=N(A\cap B^C\cap C^C)=11$