I'm currently working on an old combinatorics problem from my university, which is as follows: Let there be 9 English books, 7 French books and 10 German books. How many ways are there to choose 7 books from them, so that every language is represented?
I've found some similar asked problems but I'm still not sure what the correct answer might be...
I thought that I could handle it similar to a lottery problem, where for example the ways to get 1 out of 6 right when there are 45 numbers is: $\binom{6}{1}\binom{39}{5}$
So I came up with the answer: $\binom{7}{3}\binom{19}{4} + \binom{7}{4}\binom{19}{3} + \binom{7}{5}\binom{19}{2} + \binom{7}{6}\binom{19}{1} + \binom{7}{7}\binom{19}{0}$
I'm not very confident in the validity of this answer, can you help me?
If the books in same category are the same as one another , we can use ordinary generating functions to calculate. It is said that every book type must be represented , then
Generating function for $9$ English books where there is at least one book : $$\frac{x(1-x^9)}{1-x}=x+x^2+x^3+...+x^9$$
Generating function for $7$ French books where there is at least one book: $$\frac{x(1-x^7)}{1-x}=x+x^2+x^3+...+x^7$$
Generating function for $10$ German books where there is at least one book : $$\frac{x(1-x^{10})}{1-x}=x+x^2+x^3+...+x^{10}$$
Now , find the coefficient of $x^7$ in the expansion of them such that EXPANSION
So , the answer is $15$