I have a wording question regarding the following question:
"Four players, named A, B, C, and D, are playing a card game. A standard, well-shuffled deck of cards is dealt to the players (so each player receives a 13-card hand).
How many possibilities are there for the hand that player $A$ will get? (Within a hand, the order in which cards were received doesn't matter.)"
$\require{cancel}$ From my understanding, the possibilities for player $A$'s hand is $\binom{52}{13}$ if the cards are distributed in groups of 13s. However, if the cards are distributed one after another to each of the four players, the possibilities will be $ \cancel{ \binom{52}{1}+\binom{52-4}{1}+\binom{52-8}{1}\cdots\binom{52-48}{1} }$ correct?
Sorry I'm not familiar with how most card games are distributed among players, hence, the question. And does "within a hand" simply means all the cards in hand during the game?
Kindly advise
Updates
Thanks to Henry & JMoravitz comments,
I've changed my title and added the following:
$$ \binom{n}{k} = \binom{52}{13} = \frac{\prod_{i=0}^{n-1} \binom{52-i}{1}}{13!} $$