Probability on card suits in a standard deck Let's say we have a standard deck of $52$ cards with $13$ cards each of the standard suits of spades, clubs, diamonds, hearts. Let's say we have $p$ players and each player is dealt q cards where $pq<52$. 


*

*Without distinguishing cards by their letter (so we treat any two
cards of the same suit as the same, but two cards of different suits
as different), how many ways can these cards be distributed?

*How many ways can the cards be dealt so that someone (at least one
player) doesn't get a spade? 


You can plug in values that you find comfortable for $p$ and $q$, I just need to see work as to how it would be done so I can make it general.
I started working toward this and realized that I can just make the deck a deck of $pq$ cards and then multiply whatever formula I will get by $C(52, pq)$ to cover all of the bases. I don't know where I would go from here.
 A: Let me work out for a standard card game, e.g. bridge $(p=4,q=13)$
which should give you some idea of the process to be used.
$(1)$
Use the permutation formula with some objects identical
Total ways of distributing $ = \dfrac{52!}{13!13!13!13!} = 53,644,737,765,488,792,839,237,440,000$
$(2)$
Imagine that there are $4$ groups of $13$ slots where cards are placed.
To correspond to part $(1)$, we shall first place the spades, (which can go to at most $3$ players), and then multiply for placement of the non-spades in the remaining $39$ slots in $\dfrac{39!}{13!13!13!}$
Use PIE, principle of inclusion-exclusion
Ways to place spades to any $3$ players = $\dbinom43\dbinom{39}{13}$,
but this includes only $2$ or $1$ player having spades, double counts cases with $2$ players having spades, and if we correct for this, eliminates cases with only $1$ player having spades. Applying PIE to get the correct count, we get
$$\dbinom43\dbinom{39}{13} - \dbinom42\dbinom{26}{13} + \dbinom41\dbinom{13}{13} = 32, 427, 298, 180$$
and, of course, multiply by $\dfrac{39!}{13!13!13!}$ for placement of the non-spades. 
