Number of Non-Fiction Humourous Books for Children in a Bookstore

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The discussion revolves around calculating the number of non-fiction humorous books for children in a bookstore with a total of 54 books. The breakdown includes 28 fiction, 21 humor, and 17 children's books, with overlaps in categories such as humorous fiction and children's fiction. The solution process involves setting up an equation based on the total number of books and their classifications. The conclusion reached is that there are 2 non-fiction humorous books for children. Clarification on the interpretation of the term "fiction" in the context of the total count was also discussed.
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Homework Statement



There are 54 books in a book store: 28 are fiction, 21 are humour and 17 are children’s books. If 12 of the books are humourous fiction, 8 are children’s fiction, 3 are humourous fiction for children and 10 are serious non-fiction for adults, how many books are non-fiction humour for children.

Homework Equations



n(total) = n(C) + n(F) + n(H) – n(C + F) – n(C + H) – n(F + H) + n(F + C + H)


The Attempt at a Solution



Let x= the number of non-fiction humour for children.

n(total) = n(C) + n(F) + n(H) – n(C + F) – n(C + H) – n(F + H) + n(F + C + H)
44 = (9-x+5+3+x) + (11+9+3+5) + (9+3+x+9-x) – (5+3) – (3+x) – (3+9) + 3
44 = 46 –x
-2 = -x
x=2

I am a bit confused by the wording. Does "28 are fiction" mean that there are 28 books that are only fiction (and nothing else) or does it mean there are 28 fictional books including the humourous fiction and children’s fiction? I took it to mean the latter. Did I do this correctly? Thanks for your help.

My Diagram is attached.
 

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I think you're correct in saying that there are 28 fictional books including the humorous fiction and children's fiction since if there were 28 purely fictional stories that were nothing else you have to have more than 54 total books.
 
Looks good to me...

That's exactly how I'd solve it.
 
Thanks for your help.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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