Can't understand the answer to this combinatorics problem

Question

Consider a deck of 40 cards (10 separate cards for each of the four suits). 
Alice, Bianca, and Carlo draw from the deck respectively 5, 4, and 3 cards.
How many are the possible extractions in which Alice and Bianca have not drawn swords, while Carlo has 
extracted at least 2?

The answer is: $30\choose{5}$$\times$$25\choose{4}$$\times$$($$10\choose{2}$$\times$$21\choose{1}$+$10\choose3$$)$

But I can't understand why $10\choose3$ is inside the parentheses, something like: $30\choose{5}$$\times$$25\choose{4}$$\times$$($$10\choose{2}$$\times$$21\choose{1}$$)$+$10\choose3$ it seemed more logical to me, because I'm choosing $A+B\iff A \ or \ B \iff (exactly\ 2 \ drawn \ swords) \ or \ (at\ least \ 2\ swords)$

However it's a little different than I thought.

Answer 1

I think, in the original answer, it is basically a 3 term product. Alice's draw, times Bianca's draw, times Carlo's draw. There are Two possibilites for Carlo's draw. He either draws 2 swords and 1 other, or he draws 3 swords and 0 other. All three people draw, so their respective terms are multiplied, however, Carlo's draw, can only be one of the aforementioned possibilites, hence you have addition between them, since you're counting all the possibilites. You have to be careful with ands and ors.

Your answer reads like this:

$\bigg($Alice draws 5 cards AND Bianca draws 4 cards AND Carlo draws exactly 2 swords$\bigg)$ OR Carlo draws 3 swords.

The given answer reads like this:

Alice draws 5 cards AND Bianca draws 4 cards AND $\bigg($Carlo draws exactly 2 swords OR Carlo draws 3 swords.$\bigg)$

The main point here is, that you have to treat Carlo's draw, as ONE thing. In your answer, the last part stands on it's own, meaning Carlo draws 3 swords, and the other two doesn't do anything, hence you don't take account for a lot of outcomes, hence it is less, than the given answer. It can help, if you look at the different events as sets, like $A,B,C$ etc. and $+,\times$ as $\cup,\cap$ respectively. You applied the associative property of sets incorrectly. It only holds, if there's only one of the two set operators present.

Answer 2

Break the question down into two cases: either Carlo draws exactly $2$ swords or $3$ swords. In the former case, the number of possible extractions is ${30 \choose 5}{25\choose 4}{10\choose 2}{21\choose 1}$, because Alice must draw $5$ non-swords, Bianca must draw $4$, and Carlo must draw $1$ non-sword, and $2$ swords.

In the latter case, the number of possible extractions is ${30 \choose 5}{25\choose 4}{10\choose 3}$, by similar reasoning. Adding these two together, and factoring out the common factor gives the desired answer.