[Combinatorics]
Problem: Let $T$ = 39. Emile randomly chooses a set of $T$ cards from a standard deck of 52 cards. Given that Emile's set contains no clubs, compute the probability that his set contains three aces.

Solution: Knowing that 13 of the cards are not in Emile's set, there are $\left(\begin{array}{c}39 \\ T\end{array}\right)$ ways for him to have chosen a set of $T$ cards. Given that Emile's set contains no clubs, the suits of the three aces are fixed (i.e., diamonds, hearts, and spades). The number of possible sets of cards in which these three aces appear is therefore $\left(\begin{array}{c}36 \\ T-3\end{array}\right)$. The desired probability is therefore $\left(\begin{array}{c}36 \\ T-3\end{array}\right) /\left(\begin{array}{c}39 \\ T\end{array}\right)$. With $T=39$, this probability is $1 / 1=\mathbf{1}$, which is consistent with the fact that Emile's set contains all cards in the deck that are not clubs, hence he is guaranteed to have all three of the remaining aces.

Final answer: 1