In a competition, a school awarded medals in different categories. 36 medals in dance,12 in dramatics and 18 medals in music.If these medals went to total 45,and only 4 persons got medals in all three catogories.Using set notations, how many received in exactly two of these categories?


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Question Number 28805 by NECx last updated on 30/Jan/18

$I n a c o m p e t i t i o n, a s c h o o l a w a r d e d$ $m e d a l s i n d i f f e r e n t c a t e g o r i e s .$ $36 m e d a l s i n d a n c e, 12 i n d r a m a t i c s$ $a n d 18 m e d a l s i n m u s i c . I f t h e s e$ $m e d a l s w e n t t o t o t a l 45, a n d o n l y$ $4 p e r s o n s g o t m e d a l s i n a l l t h r e e$ $c a t o g o r i e s . U s i n g s e t n o t a t i o n s,$ $h o w m a n y r e c e i v e d i n e x a c t l y$ $t w o o f t h e s e c a t e g o r i e s ?$
	Answered by Rasheed.Sindhi last updated on 30/Jan/18

	$A : Set of medals in dance$ $B : Set of medals in drama$ $C : Set of medals in music$ $n (A) = 36, n (B) = 12, n (C) = 18$ $n (A \cup B \cup C) = 45, n (A \cap B \cap C) = 4$ $- . - . - . - . - . -$ $n (A \cup B \cup C) = n (A) + n (B) + n (C)$ $- n (A \cap B) - n (B \cap C) - n (A \cap C)$ $+ n (A \cap B \cap C)$ $n (A \cap B) + n (B \cap C) + n (A \cap C) =$ $n (A) + n (B) + n (C) - n (A \cup B \cup C)$ $+ n (A \cap B \cap C)$ $n (A \cap B) + n (B \cap C) + n (A \cap C) = 36 + 12 + 18 - 45 + 4$ $= 25$ $Each of (A \cap B), (B \cap C) & (A \cap C) contains$ $(A \cap B \cap C)$ $So,$ $Number of medals received in exactly two$ $catagaries :$ $n (A \cap B) + n (B \cap C) + n (A \cap C) - 3 \times n (A \cap B \cap C)$ $= 25 - 3 \times 4 = 13$
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