our players Blanche, Dorothy, Rose, and Sophia are playing bridge. An outcome is a particular partition of the $52$ cards into assignment of $13$ cards each to the $4$ players. Leave your answers in factorials.
- (a) Let $𝑆$ be sample space. Count the number of possible outcomes.
- (b) Let $𝐴$ be the event that each player gets an ace. How many possible outcomes are there in $𝐴$?
- (c) Let $𝐵$ (respectively $𝐷$, $𝑅$ and $𝐻$) be the events that Blanche (respectively Dorothy, Rose and Sophie) have at least one ace. Count the number of outcomes in $B\cup D\cup R\cup H$.
- (d) Count the number of possible outcomes in 𝐵 ∩ (𝐷 ∪ 𝑅 ∪ 𝐻)𝑐.
The question I really need the most help with is d, but I just want to make sure my other answers for the questions are correct as well.
For a, I got $\binom{52}{13}\binom{39}{13}\binom{26}{13}$. After choosing $13$ cards to give to a player, you must replace them leaving $n-13$ for the next player.
For b, I got $4\cdot 3\cdot 2\cdot 1 = 24$ outcomes- There are $4$ aces, and if we give $1$ to a player it leaves $3$ that the next player can get, etc.
For c, we add the possibilities that a player has $1, 2, 3,$ and $4$ aces, then multiple by the number of players.
For d, I'm not exactly sure how to go about it. How can I find the complement of possible outcomes like this? Thank you!
