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I am having trouble solving this combinatorial problem dealing with the number of different card hands possible from multiple decks of identical cards. Here is the exact question:

Use a combinatorial argument to count the number of different six-card hands that can be dealt from r combined decks, for each positive integer r.

I assumed that if r > 6 it would not effect the answer because at most a card can be repeated six times. Therefore, I interpreted the question as asking to use a combinatorial argument to count the number of different six-card hands that can be dealt from a sextuple deck.

From that point I found eleven different situations that could occur represented by the following sequences: 123456, 112345, 112234, 112233, 111234, 111223, 111222, 111123, 111122, 111112, 11111. By this notation I mean that the first sequence represents a six-card hand with zero cards repeating, the second contains 1 repeat, the third contains 2 repeats, and the fourth contains 3 repeats.

I then attempted to represent this using binomial coefficients: $$\binom{52}{6}\binom{6}{0}+\binom{52}{5}\binom{5}{1}+\binom{52}{4}\binom{4}{2}+\binom{52}{3}\binom{3}{3}+\binom{52}{4}\binom{4}{1}+\binom{52}{3}\binom{3}{1}\binom{2}{1}+\binom{52}{2}\binom{2}{2}+\binom{52}{3}\binom{3}{1}+\binom{52}{2}\binom{2}{1}\binom{1}{1}+\binom{52}{2}\binom{2}{1}+\binom{52}{1}\binom{1}{1},$$ but I was mainly confused about the sixth binomfial coefficient $\binom{52}{3}\binom{3}{1}\binom{2}{1}$ representing the sequence 111223 where is one card thrice repeated and another twice repeated. The answer I came up with is 36,288,252 possible hands.

If anyone could help verify my answer or inform me on what I am doing wrong please do.

Thank you in advance.

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  • $\begingroup$ A couple of questions. Why restrict $r$ to simply a sextuple deck? Do hands include different suits? In other words, with multiple decks, say 6, you can have 6 aces of spades or 6 aces of hearts or 3 aces of diamonds and 3 of clubs. Would those just be 6 of a kind (i assume yes, but math has to be precise...) $\endgroup$ – Eleven-Eleven Dec 5 '14 at 1:42
  • $\begingroup$ I think you need to deal with $r=1$ to $5$ also. But the answers can be obtained by removing the appropriate terms from your calculation with $6$. $\endgroup$ – André Nicolas Dec 5 '14 at 1:52
  • $\begingroup$ The way I interpreted the question is that a hand of 6 aces of spades and a hand of 6 aces of hearts are considered a different hand. Not sure if that answers your question. Does the third paragraph not explain why I chose to restrict r to a sextuple deck? $\endgroup$ – Connor Berg Dec 5 '14 at 2:10
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Your numerical answer is correct, but this (the $r=6$ case) can be dealt with much more efficiently using balls and urns. The answer should be a single binomial coefficient.

However, Eleven-Eleven and André are correct that you should not assume $r=6$, especially given the wording of the problem. For the general case, you can count the complement in order to simplify things, though some mild casework may be unavoidable.

It may be instructive to test out your answer for very small values of $6$ and $52$.

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  • $\begingroup$ Okay, so using the ball-and-urn technique I can simplify my above expression to $\binom{52+6-1}{6}$. I am confused, however, on the complement counting technique. $\endgroup$ – Connor Berg Dec 5 '14 at 2:23
  • $\begingroup$ @ConnorBerg For example, take $r=5$. Of the 36,288,252 hands, how many require six decks? $\endgroup$ – Slade Dec 5 '14 at 2:27
  • $\begingroup$ Wouldn't only 52 hands require six decks? $\endgroup$ – Connor Berg Dec 5 '14 at 2:31
  • $\begingroup$ @ConnorBerg Correct, which takes care of $r=5$. For $r=4$, ask how many require five decks. $r=3$ may be the most difficult case. $\endgroup$ – Slade Dec 5 '14 at 2:59
  • $\begingroup$ Here is what I came up with: 52 hands require six decks, 2,652 hands require five decks, 68,952 hands require four decks, 1,151,852 hands require three decks, and 14,641,250 hands require two decks. $\endgroup$ – Connor Berg Dec 5 '14 at 3:17

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