combinatorics: deck of cards and suits

Question

I feel confused about selecting suits and cards that each suit corresponds to.

Suppose we have a standard deck of cards, and we want to form a 6 hands with at most 3 suits.

I understand this needs to be spliced into 3 cases to discuss, which are 1 suit case, 2 suits case, and 3 suits case.

But I feel confused when it comes to select cards.

For example, I am not sure if I understand this correctly, but I feel like if we only need to form cards with a single suit, then this can be done by $4(13 * 12 * 11 * 10 * 9 * 8)$ ways, since each suit has 13 cards, and we have 4 suits here.

And when it comes to the 2 suits case, I feel like we only need to draw from the total of 26 cards to form such hands, and there are 6 ways to do this, but I am not sure if my understanding about the 26 cards is right or what.

And for the 3 suits case, I feel like it's just choose 3 suits out of 4, and draw cards from total of 39 cards.

Is my understanding towards the total number of cards in each case correct? If not, why so?

Answer 1

There is no need to approach by breaking into cases. Case work will just become too messy and does not generalize nicely to other sized hands. Consider answering the question instead as: "How many ways are there to form a six-card hand such that at least one of the suits is missing?"

Let $A$ be the event that clubs are missing, $B$ that diamonds are missing, $C$ that hearts are missing, and $D$ that spades are missing respectively. You are tasked with finding $\Pr(A\cup B\cup C\cup D)$

This expands by inclusion exclusion to be:

$$\Pr(A\cup B\cup C\cup D) = \Pr(A)+\Pr(B)+\Pr(C)+\Pr(D)-\Pr(A\cap B)-\Pr(A\cap C)-\dots - \Pr(C\cap D) + \Pr(A\cap B\cap C)+\dots + \Pr(B\cap C\cap D) - \Pr(A\cap B\cap C\cap D)$$

By symmetry, this simplifies to be:

$$ = 4\cdot \Pr(A) - 6\cdot \Pr(A\cap B) + 4\cdot \Pr(A\cap B\cap C)$$

noting that we could leave off the intersection of all four since it is impossible for all suits to be missing simultaneously.

The probability that clubs are missing? That is answered simply by a hypergeometric distribution argument and is $\Pr(A) = \dfrac{\binom{39}{6}}{\binom{52}{6}}$. Similarly for the probability that both clubs and diamonds are missing as $\dfrac{\binom{26}{6}}{\binom{52}{6}}$ and so on.

Your probability is then:

$$\dfrac{4\cdot\binom{39}{6} - 6\cdot\binom{26}{6}+4\cdot\binom{13}{6}}{\binom{52}{6}}$$

Answer 2

There are two ways of enumerating the number of possible $(6)$ card hands, from a $(52)$ card deck, when the cards are drawn without replacement, and where you want to reject any $(6)$ card hands that involve all 4 suits.

One way is to enumerate the number of $(6)$ card hands that do involve all 4 suits, and deduct that from

$$\binom{52}{6} = \frac{(52)!}{(6!)[(52-6)!]}.$$

The alternative method is the direct approach, which is detailed below. I have chosen this approach because, while it is more difficult, it is also more educational.

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You have a $(3)$ step process.

First, you have to identify all of the mutually exclusive ways of partitioning the number $(6)$ into the sum of three non-negative integers.

These are

• 6,0,0
• 5,1,0
• 4,2,0
• 4,1,1
• 3,3,0
• 3,2,1
• 2,2,2

Then, one at a time, you have to enumerate each of the three mutually exclusive patterns above. In such an enumeration, the second step is to enumerate the total number of different ways of assigning suits, to fit the pattern. I refer to this as computing the suit-factor.

Then, the third step is to enumerate the number of ways of choosing the cards within each of the pertinent suits. I refer to this as computing the combination-factor.

Throughout this discussion, you need to know that if $~n \in \Bbb{Z^+},~$ and $~k \in \{0,1,2,\cdots,n\},~$ that the number of ways of choosing $k$ things out of $n$ things, sampling without replacement, where order of selection is regarded as irrelevant is

$$\binom{n}{k} = \frac{n!}{k![(n-k)!]}.$$

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$\underline{\text{Case 1: Pattern = 6-0-0}}$

There are $(4)$ ways of choosing which suit to draw the cards from. So, you start with a suit-factor of $(6)$.

Then, you have the combination-factor, that represents the number of ways of selecting the cards, once the suit assignments are made. There are $~\displaystyle \binom{13}{6}~$ ways of choosing the $(6)$ cards within one suit.

Therefore, the overall enumeration here is

$$4 \times \binom{13}{6}. \tag1 $$

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$\underline{\text{Case 2: Pattern = 5-1-0}}$

There are $(4)$ ways of choosing which suit to select $(5)$ cards from. Once this is done, there are then $(3)$ ways of selecting which suit to draw the single card.

So, the suit-factor is $(4 \times 3 = 12).$

Then, the combination-factor is $~\displaystyle \binom{13}{5} \times \binom{13}{1}.$

Therefore, the overall enumeration here is

$$12 \times \binom{13}{5} \times \binom{13}{1}. \tag2 $$

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$\underline{\text{Case 3: Pattern = 4-2-0}}$

There are $(4)$ ways of choosing which suit to select $(4)$ cards from. Once this is done, there are then $(3)$ ways of selecting which suit to select $(2)$ cards from.

So, the suit-factor is $(4 \times 3 = 12).$

Then, the combination-factor is $~\displaystyle \binom{13}{4} \times \binom{13}{2}.$

Therefore, the overall enumeration here is

$$12 \times \binom{13}{4} \times \binom{13}{2}. \tag3 $$

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$\underline{\text{Case 4: Pattern = 4-1-1}}$

There are $(4)$ ways of choosing which suit to select $(4)$ cards from. Once this is done, there are then $(3)$ ways of selecting which suit will be unused.

So, the suit-factor is $(4 \times 3 = 12).$

Then, the combination-factor is $~\displaystyle \binom{13}{4} \times \binom{13}{1} \times \binom{13}{1}.$

Therefore, the overall enumeration here is

$$12 \times \binom{13}{4} \times \binom{13}{1} \times \binom{13}{1}. \tag4 $$

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$\underline{\text{Case 5: Pattern = 3-3-0}}$

The suit-factor is $~\displaystyle \binom{4}{2} = 6.$

The combination-factor is $~\displaystyle \binom{13}{3} \times \binom{13}{3}.$

Therefore, the overall enumeration here is

$$6 \times \binom{13}{3} \times \binom{13}{3}. \tag5 $$

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$\underline{\text{Case 6: Pattern = 3-2-1}}$

The suit-factor is $~\displaystyle 4 \times 3 \times 2 = 24.$

The combination-factor is $~\displaystyle \binom{13}{3} \times \binom{13}{2} \times \binom{13}{1}.$

Therefore, the overall enumeration here is

$$24 \times \binom{13}{3} \times \binom{13}{2} \times \binom{13}{1}. \tag6 $$

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$\underline{\text{Case 7: Pattern = 2-2-2}}$

There are $(4)$ ways of selecting the suit to be unused. Therefore, the suit-factor is $4.$

The combination-factor is $~\displaystyle \binom{13}{2} \times \binom{13}{2} \times \binom{13}{2}.$

Therefore, the overall enumeration here is

$$4 \times \binom{13}{2} \times \binom{13}{2} \times \binom{13}{2}. \tag7 $$

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$\underline{\text{Final Computation}}$

To get the overall enumeration, all that you have to do is use a hand calculator against the computations in each of
(1), (2), (3), (4), (5), (6), and (7), above.

Then, take the overall sum of these seven computations.