Question 1055594
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From the elementary set theory, there is the formula


n(A&#8746;B&#8746;C) = n(A) + n(B) + n(C) - n(A&#8745;B) - n(A&#8745;C) - n(B&#8745;C) + n(A&#8745;B&#8745;C).   (1)


or, equivalently, 

n(A&#8745;B&#8745;C) = n(A&#8746;B&#8746;C) - n(A) - n(B) - n(C) + n(A&#8745;B) + n(A&#8745;C) + n(B&#8745;C).   (2)


Therefore,  n(A&#8745;B&#8745;C) = 30 - 15 - 18 - 16 + 7 + 6 + 7 = 1. 


Regarding the explanations, read the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/misc/Advanced-probs-counting-elements-in-sub-sets-of-a-given-finite-set.lesson>Advanced problems on counting elements in sub-sets of a given finite set</A>

in this site.


Everything is explained there.
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Also, you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lesson is the part of this online textbook under the topic "<U>Miscellaneous word problems</U>".