# Combinatorics of card game states

Suppose we have a card game with some $n$ cards and $m$ players, where $m \mid n$. Each player starts with $\frac{n}{m}$ cards.

How many starting states does the game have?

How many states can the game be in after $k$ rounds, if in every round, every player drops $j$ of their remaining cards at random?

• Does m divide n? – Peter Jan 29 '14 at 20:06
• @Peter, yes, and I've complemented the question with the information. – Nerius Jan 29 '14 at 20:09

If $l = \frac{n}{m}$ , we have

$$\frac{n!}{(l!)^m}$$

starting states.

After $k$ rounds, every player has $l-kj$ cards left.

So, the total number of cards is $m(l-kj) = n-mkj$

So, the number of states is now

$$\binom{n}{n-mkj}\frac{(n-mkj)!}{(l-kj)!^m}$$

• Thanks, however this answer assumes that every round the player disposes of 1 card, whereas they dispose of some j. – Nerius Jan 29 '14 at 20:20