# Counting number of ways of splitting a card deck

Suppose that we have a $52$-card deck. We are interested to find how many different combinations there could be if we divide this $52$-card deck in two parts, so that in each part there are $2$ aces.

What I am thinking is that we have $4$ aces, we can choose two of them in $\dfrac{4!}{2!}=12$ ways.

Also $52$ cards, we can divide in two parts in $\dfrac{52!}{2!}$ ways, so I have to multiply $12$ by $\dfrac{52!}{2!}$? It is a huge number, so am I missing something?

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Note that sometimes it is acceptable to give an answer in terms of factorials and combinatorial numbers, so you might not actually have to perform the multiplication. – talmid Jun 24 '12 at 8:17
If you are splitting the deck between Alice and Bob, the answer is $6\times 2^{48}$. If you are just splitting, the answer is $3\times 2^{48}$. – André Nicolas Jun 24 '12 at 17:09

## 3 Answers

There are 3 ways to split the aces: SH vs DC, SD vs HC, and SC vs HD (that's Spades, Hearts, Diamonds, Clubs).

Having split the aces, each of the other 48 cards has 2 possible destinations. So the answer would be $$3\times2^{48}$$

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sorry one question,why is here three way? – dato datuashvili Jun 24 '12 at 8:51
aa i understood right about aces,but about 48 card,would not it be $48!/2!$? – dato datuashvili Jun 24 '12 at 8:55
Suppose the deck had only 5 cards. Would the part of the answer for the 1 non-ace be $1!/2!$? or would it be $2^1$? – Gerry Myerson Jun 24 '12 at 9:07
right thanks very much – dato datuashvili Jun 24 '12 at 9:09
As @Gerry Myerson said it changes if you distinguish one hand from the other. So the correct answer is either 3 * 2^48 or 6*2^48. In the case of 5 cards, the two possible answers are either 3* 2^1 or 6* 2^1 – jeff Jun 24 '12 at 9:23

There are $\binom{4}{2}$ possibilities for picking 2 aces out of 4 and $\binom{48}{24}$ possibilities for picking 24 out of 48 non-aces. So the answer is $\binom{48}{24} \binom{4}{2}=193485622098600$.

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But it didn't say the two parts had to have 26 cards each. – Gerry Myerson Jun 24 '12 at 8:10
so is it possible to express answer like this?$48!/2!$*$4!/2!$? – dato datuashvili Jun 24 '12 at 8:27

Take out the 4 Aces. Hand #1 gets 2 Aces. Hand #2 gets 2 Aces.

The 4 Aces Spades, Clubs, Hearts , Diamonds (SCHD)

Hand #1 possiblities SC, SH, SD, CH, CD, HD for 6 possibilities

The remaining 48 cards each has a 50% chance of going to either hand. 2^48

The total possible hands = 6*2^48

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That's if you distinguish Hand 1 from Hand 2. The question just said how many combinations if you divide the deck in two parts, so I interpret that to mean you don't distinguish the two hands. But it's a matter of interpretation, and only OP knows for sure. – Gerry Myerson Jun 24 '12 at 9:09