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I have $52$ cards from $4$ series ($13 \times 4=52 \text{ cards}$). Each series are numbered from $1$ to $13$. In how many possible ways you can draw from the deck $6$ cards, so that all your six cards has two numbers exactly?

My answer:

First card - $\binom{13}{1} \cdot \binom{8}{4}$ (choosing from the $8$ possibilities, $4$ places I can put my cards. choosing $4$ because I can't have more than $4$ cards of the same number)

Second card - $\binom{13}{1} \cdot \binom{7}{4}$ (same, but $7$)

Total - First card + Second card

Not quite sure about my answer. I am pretty sure I am counting more than there is. Help will be much appreciated.

2 Answers 2

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The number of different ways to obtain $2$ out of $13$ different numbers is \begin{align*} \binom{13}{2} \end{align*}

Next we consider all valid configurations which can be obtained with two numbers.

Since there are always four cards with the same number and we take six cards from the deck there are three types of valid configurations

  • $(4,2)$: Four cards with the first number and two cards with the second number. There are \begin{align*} \frac{6!}{4!2!}=\binom{6}{2} \end{align*} different configurations of this type.

  • $(3,3)$: Three cards with the first number and three cards with the second number , giving \begin{align*} \frac{6!}{3!3!}=\binom{6}{3} \end{align*} different configurations.

  • $(2,4)$: Two cards with the first number and four cards with the second number, giving $$\frac{6!}{2!4!}=\binom{6}{2}$$ different configuration as in the first case.

We conclude:

There are \begin{align*} \binom{13}{2}\left(2\binom{6}{2}+\binom{6}{3}\right)=3900 \end{align*} different configurations to draw six cards with two numbers from the deck.

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    Maybe I'm wrong here or miss something, but the OP question is about the number of possibilities to draw six cards that has two numbers exactly. – 2017-01-10
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    @SalechAlhasov: Thanks for the hint. Wording corrected. – 2017-01-11
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If I am correctly understanding this to be asking "how many ways can you draw 6 cards from a standard 52 card deck such that no more than two unique values are represented in the hand" then the answer is \begin{align*} \binom{13}{2}*\binom{8}{2} = 2184 \end{align*} Explanation: \begin{align*} \binom{13}{2}\end{align*} selects which two values the hand will contain. \begin{align*} \binom{8}{2}\end{align*} chooses which of the 8 cards that have those two values will actually be in the hand. This is assuming that we do not care about the order of the cards in the hand.

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    I believe your way is under-counting. Well picking the six cards have a few cases. For example, let's say I pick number 2 and 3. I can have four 2's and two 3's, two 2's and four 3's, or three 2's and three 3's. – 2017-01-10
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    All of those cases are accounted for by the (8 choose 6). In your example the 8 cards being chosen from are 2H, 2C, 2S, 2D, 3H, 3C, 3S, 3D. Choosing 6 of them could give us 2H, 2C, 2S, 2D, 3H, 3C (4 twos, 2 threes) or 2S, 2D, 3H, 3C, 3S, 3D (2 twos, 4 threes) or any other combination. – 2017-01-11
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    It is possible that I am misunderstanding something, but Markus's answer appears incorrect to me. For example, you say there are 6 choose 2 = 15 ways to select 4 of the first number and 2 of the second, but there are only 6: you have all 4 of the first number, and your choice of any 2 of the second, giving 4 choose 2 = 6. – 2017-01-11
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    @Rance: The suits are not considered in the problem. So, we don't consider $2H, 2C, 2S, 2D$, but four two's $2222$ instead. I think there are $\binom{6}{2}=15$ different configurations with four $1$'s and two $2$'s on the deck: $111122,111212,111221,112112,112121,112211,121112,121121,121211,122111,211112,211121,211211,212111,221111$ – 2017-01-11
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    @Markus: Ok, I understand what you are saying. Your answer is correct if the order of the draws matters and the specific cards drawn does not. I understood the question to be asking how many unique 6-card hands meeting those conditions could be drawn from the deck (the question you would need to answer if you wanted to then calculate the probability of drawing such a hand). – 2017-01-11