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## Here's the Solution to this Question

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

$11+10+6+4+6+5+3=45$ $N(\text{chemistry or biology only})=N(C\cap B\cap A^C)=5$

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