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I'm an undergraduate taking my first-ever upper division math class. It's Combinatorics, which is pretty

There are $\binom{52}{5}$ different hands possible in a single poker deck. In a double deck, however, there are more than that, because some cards can be repeated.

The way that I've been attempting to enumerate this is by going through the different possibilities,

  • abcde
  • aabcd
  • aabbc

and trying to enumerate the number of different combinations of cards that can be rearranged to those specific permutations.

So, for example, aa-bcd has $\binom{52}{1}$ possibilities for the "aa" pair, and a further $\binom{52-1}{3}$ possibilities for the "bcd" part of the hand. So would there be $$\binom{52}{1} \binom{51}{3}$$ possiblities?

By this logic, I come up with the summation

$$\binom{52}{5} + \binom{52}{1}\binom{51}{3} + \binom{52}{2}\binom{50}{1}$$,

(Edit: Slight mistake, the last one should be $\binom{50}{1}$ not $51$. Pointed out by Long, below.)

but since I can't find a hard answer, I'm finding it difficult to double-check my work.

What do you think? Is this the right way to "think through" the problem? Have I made an important mistake (likely)?

Edit: Thank you for all of your help. I took the confirmation and generating function technique I learned here to write an extension to this, on how to count the number of 5 card hands from a double, triple, and quadruple deck.

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  • $\begingroup$ You should get $3\,748\,160$ if you make the correction as suggested by Kevin Long's answer. Another way to look at it is using the generating function $(1+x+x^2)^{52}$ then taking the coefficient of $x^{5}$. With a bit of algebraic manipulation this gives $\binom{56}{5}-\binom{52}{1}\binom{53}{2}$. $\endgroup$ – N. Shales Feb 1 '18 at 18:31
  • $\begingroup$ Yes, now you have edited your answer is correct! $\endgroup$ – N. Shales Feb 1 '18 at 19:27
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Your reasoning is correct. The different cases are just counting how many pairs of repeated cards you have (either one, two, or three). You should just note that you've made a slight mistake on the third case- see if you can spot what it is.

EDIT: With the edit, your answer is correct.

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  • $\begingroup$ Thank you for the confirmation and mistake-catching. $\endgroup$ – Andrew Quinn Feb 2 '18 at 1:31
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We have that for your hand, you could have any particular card appear in it either no times, once, or twice, so the solution can be found as the coefficient of $x^5$ of the expanded function $$f(x) =(1+x+x^2)^{52}$$

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  • $\begingroup$ Very cool! Generating functions are an incredible tool when it comes to things like this. $\endgroup$ – Andrew Quinn Feb 2 '18 at 1:30

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