Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

3 Answers

up vote 2 down vote accepted

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 $$3\times2^{48}$$

share|improve this answer
    
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
1  
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
add comment

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 $\binom{48}{24} \binom{4}{2}=193485622098600$.

share|improve this answer
1  
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
add comment

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

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.