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Question : In a school, the students are fans of one or more of the four actors- A,B,C and D.The four actors given in the above order are liked by $230,180,180$ and $220$ students respectively.The no. of students who like exactly $2$ actors for any two actors is $20$ (for each two).There are $30$ students who like all the four actors but there is nobody who likes exactly three out of four actors.

-How many students are there in school?

Answer :

$N(A \cup B \cup C \cup D) = N(A) + N(B) + N(C) + N(D) - N(A \cap B) - N(A \cap C) - N(A \cap D) - N(B \cap C) - N(B \cap D) - N(C \cap D) + N(A \cap B \cap C) + N(A \cap B \cap D) + N(A \cap C \cap D) + N(B \cap C \cap D) - N(A \cap B \cap C \cap D)=$

$180+180+220+230+120-30=660.$

But answer is supposed to be $600$.

Jean Marie
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dhruv
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  • The key words are "like exactly 2 actors". You have applied a formula that would work if the wording were "like at least those 2 actors". (And then it would be impossible to have a non-empty subset who like all four actors without having subsets who like at least three specified actors.) – David K Jul 22 '23 at 06:22

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When summing the students who like each actor ($N(A) + N(B) + N(C) + N(D) = 230 + 180 + 180 + 220$), the $30$ students who like all four actors are counted four times; i.e. these $30$ are overcounted by $4-1=3$ times. So you have to subtract three times their contribution: $$230 + 180 + 180 + 220 - (2-1) \times 120 - (4-1) \times 30 = 600$$

Tony Mathew
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