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 $pq<52$.

  1. 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?
  2. 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.

  • 1
    $\begingroup$ I'm afraid this is going to be a horrible mess (unless $p$ and $q$ are chosen unrealistically to make it manageable). $\endgroup$
    – joriki
    Jun 2 '16 at 21:40
  • $\begingroup$ @joriki yeah you can choose like 5 and 6 for p and q. I just want to see how someone would work through this to get an idea of the methods to use here. It's not about the particular answer, just the process by which I could reach an answer so I can make a general formula. I need it for some work I am doing. $\endgroup$
    – edupppz
    Jun 2 '16 at 21:42

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.


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$


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.

  • $\begingroup$ I don't understand the reasoning behind the second part. I understand that in the first we sort of have a sequence with repetition where we have something like dividers which determine who gets what hand. $\endgroup$
    – edupppz
    Jun 3 '16 at 15:42
  • $\begingroup$ I have elaborated and homogenised the two parts. My previous comment got deleted inadvertently, I'd think it isn't needed again. $\endgroup$ Jun 4 '16 at 7:07
  • $\begingroup$ I am sorry I haven't responded more promptly, I am traveling the country and have very sporadic access to the internet. I worked through the problem and I think your answer to part 1 poses an issue. If you are handed a spade and then a club for a hand of two and another time you are dealt a club and then a spade, your method counts those deals differently. I figured out how to bypass that. So for p,q as given above, we just multiply choices: C (52,q) C (52-q,q)... C (52-(p-1)q,q). Here we only differentiate between the separate kinds of hands dealt, so it is nice. $\endgroup$
    – edupppz
    Jun 6 '16 at 14:29
  • $\begingroup$ Now I am working on solving how if we have a set of size K and take p sets of size q, _pq_≤ K, how many ways can we pick sets such that one set has no members of an arbitrarily chosen subset of size L. This is just the general form of the second part. I think I understand the PIE, but I'll look at it more closely to make sure. $\endgroup$
    – edupppz
    Jun 6 '16 at 14:42
  • $\begingroup$ Your comment re double counting in part $1$ is not correct. Pl. note that $\binom{52}{13,13,13,13} = \frac{52!}{13!13!13!13!} = \binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}$ $\endgroup$ Jun 6 '16 at 15:42

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