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In a group of 100, 50 are having an umbrella, 60 have a hat and 80 have sunglasses. 70 are such that they don’t have both – an umbrella and a hat. Similarly 50 are such that they do not have both hat and sunglasses and 60 do not have both umbrellas and sunglasses. If there are 5 who do not own any of the three items, find the number of people who own all three items.

I couldn't solve the question using 2 equations those who wear exactly one of the three -n (A ) + n (B ) + n (C ) - 2 (n (A ∩ B ) + n (B ∩ C ) + n (A ∩ C )) + 3 × n (A ∩ B ∩ C ) and the general formula to find the union - n (A ) + n (B ) + n (C ) - n (A ∩ B ) - n (B ∩ C ) - n (A ∩ C ) .

I took the sum of 70 , 50 , 60 as the sum of those who wear exactly one of the three and 95 as the union

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Let $A$ be the set of people who own an umbrella, $B$ be the set of people who own a hat, and $C$ be the set of people who own sunglasses. We want to compute the number of people who own all three items, which is equal to $|A \cap B \cap C|.$

The Principle of Inclusion-Exclusion asserts

$$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |A \cap C| + |A\cap B\cap C|. $$

Here, $|A| = 50, |B| = 60, $ and $|C| = 80$. Also, we know there are $70$ people who are not in the set $A \cap B$, which means that there are $100 - 70 = 30$ people in this set. Thus, $|A \cap B| = 30$. Using similar reasoning, we can find $|B \cap C| = 50$ and $|A \cap C| = 40$. Also, we know $|A \cup B \cup C| = 5$ for the reason you described. Therefore,

$$|A| + |B| + |C| - |A \cap B| - |B \cap C| - |A \cap C| + |A\cap B\cap C| $$

$$5= 50 + 60 + 80 - 30 - 50 - 40 - |A \cap B \cap C| $$

$$\implies |A \cap B \cap C| = \boxed{65} $$

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  • $\begingroup$ Well the answer given for this particular problem is 25 $\endgroup$ – yash471994 Nov 27 '18 at 6:37

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