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In a game of Skat there are 4 suits (spades, hearts, diamonds, clubs) and 8 values (7, 8, 9, 10, jack, queen, king, ace) yielding 32 cards altogether. I'm trying to figure out in how many ways $k \geq 4$ cards can be picked, such that every suit is represented. This is my approach:

Pick the first 4 cards. For this there are $8^4$ possibilities (8 choices from from each suit).

For the remaining $k-4$ cards to be picked there are $28 \choose k-4$ possibilities.

The order, in which the cards were picked doesn't matter, so divide by $k!$.

In total: $\frac{8^4 \cdot {28 \choose k-4}}{k!}$

Is this correct or am I missing something?

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  • $\begingroup$ What do you mean by "color"??? $\endgroup$ – barak manos Jul 15 '16 at 12:42
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    $\begingroup$ Suits = colors; the are used interchangeably $\endgroup$ – Aganju Jul 15 '16 at 12:43
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    $\begingroup$ You are missing something. I believe you are going to need inclusion-exclusion to solve this problem. You have counted some choices several different ways. Say $k=5$ and you want to get $AK$ of spades and the ace in each other suit. You've counted that case twice, because you could choose $A,A,A,A$ in your first step, then $K,A,A,A$ another time. $\endgroup$ – Thomas Andrews Jul 15 '16 at 12:49
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Use inclusion/exclusion principle:

  • Include the number of ways to choose cards from at most $\color\red4$ suits: $\binom{4}{\color\red4}\cdot\binom{8\cdot\color\red4}{k}$
  • Exclude the number of ways to choose cards from at most $\color\red3$ suits: $\binom{4}{\color\red3}\cdot\binom{8\cdot\color\red3}{k}$
  • Include the number of ways to choose cards from at most $\color\red2$ suits: $\binom{4}{\color\red2}\cdot\binom{8\cdot\color\red2}{k}$
  • Exclude the number of ways to choose cards from at most $\color\red1$ suits: $\binom{4}{\color\red1}\cdot\binom{8\cdot\color\red1}{k}$

Hence the total number of ways is:

$$\sum\limits_{n=0}^{3}(-1)^{n}\cdot\binom{4}{4-n}\cdot\binom{8\cdot(4-n)}{k}$$

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  • $\begingroup$ While your overall result is correct, the formulas for 2 and 3 suits are incorrect. For example for both, 2 and 3, you triple count the number of ways to draw 1 color only. (( 4 chose 2) yields the cases AB, AC, AD which all contain color A. Now, (16 choose 4) includes the cases where you choose from color A only.) $\endgroup$ – Nils Oct 16 '19 at 15:53

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