Choosing a combination of books, under given restrictions. 
Mary has on her bookshelf 5 novels, 5 biographies, and 8 textbooks. Mary decides to take three novels and four non-fiction books with at least one of the non-fiction books a biography. How many ways are there to make such a selection?

I thought that the answer would be obtained using multiplication of several combinations, as there are multiple object types.
My working was:
Three novels are selected from five, and
$$^5C_3 = {5 \choose 3} = 10$$
In terms of non-fiction, we have biographies and textbooks, four must be selected. One of these four must be chosen from the five biographies, so we have
$$^5C_1 = {5 \choose 1} = 5$$
The remaining three non-fiction choices can be either biographies or textbooks, one biography has already been selected so there are only twelve non-fiction books remaining, so the combination is
$$^{12}C_3 = {12 \choose 3} = 220$$
Using the multiplication principle, the number of combinations of non-fiction books is simply $5 \times 220 = 1100$.
Again using this principle, the number of combinations for a selection of 3 novels and four non-fiction books containing at least one biography must be $10 \times 1100 = 11000.$
I entered this answer and it was marked incorrect. Any help would be appreciated.
 A: As you noted, there are $\binom{5}{3}$ ways to choose the fiction books. 
There are $13$ non-fiction books. If there were no restrictions, there would be $\binom{13}{4}$ ways to choose $4$ of them. However, all $4$ being textbooks is forbidden. So there are $\binom{8}{4}$ forbidden non-fiction choices.
Thus there are $\binom{13}{4}-\binom{8}{4}$ allowed non-fiction choices, for a total of $\binom{5}{3}\left(\binom{13}{4}-\binom{8}{4}\right)$.
Remark: The problem with your approach is that you are doing multiple-counting. You chose "a" biography, and then $3$ books from the remaining $12$. Suppose that your "first" choice is biography X, and that among your $\binom{12}{3}$ choices, you chose biography Y. You counted this as different from choosing biography Y as your first choice, and biography X anong your second choices. 
A: In case the above answer wasn't intuitive to you (it wasn't to me, for some reason), there's also another way to solve the problem.  It's a lengthier calculation, but if you're not super pressed for time, it is possible that it'll make more sense to you (if you happen to think the way I do).

I started by adding up the 4 ways to select the biographies & textbooks (Bs/TBs).

There are 4 cases:

 · 1 B / 3 TBs 
 · 2 Bs / 2 TBs 
 · 3 Bs / 1 TB 
 · 4 Bs
Add them up...
${5 \choose 1}{8 \choose 3} + {5 \choose 2}{8 \choose 2} + {5 \choose 3}{8 \choose 1} + {5 \choose 4} = 645$ 
And then multiply times the ways to pick the novels, ${5 \choose 3}=10$ 
$645 \times 10 = 6450$ 
I know this feels clunkier, but even after looking at the above subtraction method, I can tell that's unlikely to be the way my brain frames the solution to similar problems in the future, and maybe if this is true for someone else, my solution will be a helpful addition (no pun intended) to the prior solution.
