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In a school, all the students play at least one of three indoor games - chess, carrom and table tennis, 60 play chess, 50 play table tennis, 48 play carrom, 12 play chess and carrom, 15 play carrom and table tennis, 20 play table tennis and chess.
What can be the minimum number of students in the school?
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What can be the minimum number of students in the school?
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The correct answer is:
111
Number of students who play chess, $\mathrm{n}(\mathrm{A})=60$ Number of students who play tennis, $\mathrm{n}(\mathrm{B})=50$ Number of students who play carrom, $\mathrm{n}(\mathrm{C})=48$ Given, $n(A \cap B)=20$
$n(B \cap C)=15$
$n(A \cap C)=12$
$\mathrm{n}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})+\mathrm{n}(\mathrm{C})-\mathrm{n}(\mathrm{A} \cap \mathrm{B})-\mathrm{n}(\mathrm{B} \cap$
C) $-n(A \cap C)+n(A \cap B \cap C)$
$=60+50+48-20-15-12+n(A \cap B \cap C)$
$=111+\mathrm{n}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})$
So, minimum number of students $=111$
$n(B \cap C)=15$
$n(A \cap C)=12$
$\mathrm{n}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})+\mathrm{n}(\mathrm{C})-\mathrm{n}(\mathrm{A} \cap \mathrm{B})-\mathrm{n}(\mathrm{B} \cap$
C) $-n(A \cap C)+n(A \cap B \cap C)$
$=60+50+48-20-15-12+n(A \cap B \cap C)$
$=111+\mathrm{n}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})$
So, minimum number of students $=111$
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