9. Alberto, Barbara, Chiara, and Davide shuffle a deck of 40 cards and then distribute 10 to each. Alberto looks at his and exclaims: "How strange, I don't have any spades." Knowing this information, what is the probability that Barbara also has no spades? (Note: the spades are 10).
(A) $\frac{30 !}{20 ! 40 !}$
(B) $\frac{20 !}{10 ! 30 !}$
(C) $\frac{30 ! 30 !}{20 ! 40 !}$
(D) $\frac{20 ! 20 !}{10 ! 30 !}$
(E) $\frac{30 ! 10 !}{40 !}$
9. The answer is $(\mathbf{D})$. The starting point is the definition of probability as

$$
\frac{\text { favorable cases }}{\text { possible cases }} \text {. }
$$

We ignore Alberto's cards; the situation simplifies: now we know there are 30 cards, 10 of which are spades, and we want to determine the probability that none of the 10 cards in Barbara's hand are spades. Suppose the hands of the three players are not sets of ten cards but ordered sets of 10 cards: this does not change the probabilities because it is equivalent to multiplying by 10! 10! 10! (the possible re-orderings of the three hands) both the number of favorable cases and the number of possible cases. Now, the possible cases are clearly 30! (we choose one by one a card to occupy one of the 30 distinct "positions" available...). The favorable cases can be counted in this way: first, we choose one by one the cards for Barbara, knowing that they must be chosen from the 20 non-spades, and we can do this in $20 \cdot 19 \cdots 11=\frac{20 !}{10 !}$ different ways; then we choose the arrangement of the remaining 20 cards, without further constraints, in one of the 20! possible ways. The two choices are clearly independent; therefore, the favorable cases are $\frac{20 ! 20 !}{10 !}$. The probability, calculated based on (1), is thus $\frac{20 ! 20 !}{10 ! 30 !}$.