Operations on Sets
Question

In a class of 30 pupils, 12 take needle work, 16 take physics and 18 take history. If all the 30 students take at least one subject and no one takes all three then the number of pupils taking 2 subjects is

Easy
Solution

Given From, $\mathrm{n}\left(\mathrm{N}\cup \mathrm{P}\cup \mathrm{H}\right)=\mathrm{n}\left(\mathrm{N}\right)+\mathrm{n}\left(\mathrm{P}\right)+\mathrm{n}\left(\mathrm{H}\right)-\mathrm{n}\left(\mathrm{N}\cap \mathrm{P}\right)$$-\mathrm{n}\left(\mathrm{P}\cap \mathrm{H}\right)-\mathrm{n}\left(\mathrm{N}\cap \mathrm{H}\right)+\mathrm{n}\left(\mathrm{N}\cap \mathrm{P}\cap \mathrm{H}\right)$$\therefore \mathrm{n}\left(\mathrm{N}\cap \mathrm{P}\right)+\mathrm{n}\left(\mathrm{P}\cap \mathrm{H}\right)+\mathrm{n}\left(\mathrm{N}\cap \mathrm{H}\right)=16$Now, number of pupils taking two subjects$=\mathrm{n}\left(\mathrm{N}\cap \mathrm{P}\right)+\mathrm{n}\left(\mathrm{P}\cap \mathrm{H}\right)+\mathrm{n}\left(\mathrm{N}\cap \mathrm{H}\right)-3\mathrm{n}\left(\mathrm{N}\cap \mathrm{P}\cap \mathrm{H}\right)$$=16-0=16.$

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