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In a deck of cards there are $52$ different cards. Out of which $4$ jacks, $4$ queens, $4$ kings and $4$ aces. Assuming a valid hand is composed by any $13$ cards. In how many different hands there are at least one from each of the following: Jack, Queen, King and Ace?

My try. Using complement:

All possible hands with no restrictions: $\binom{52}{13}$.

All possible hands without jacks, queens, kings and aces: $\binom{36}{13}$.

Hence we have that the desired hands are $\binom{52}{13}-\binom{36}{13}$.

However, that doesn't correspond with the final solution in my book. Can anyone explain what am I doing wrong and how it should be solved?

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It says "at least one from EACH of Jack, Queen, King and Ace", so a hand is not acceptable if it does not contain a Jack, OR a Queen, OR a King and, OR an Ace. Let $J$ be the set of hands without Jacks, and let $Q$, $K$, and $A$ be similarly defined. Then the number of such hands is $$\binom{52}{13}-|J\cup Q\cup K\cup A|$$ that is, by using the Inclusion–exclusion principle, $$\binom{52}{13}-\binom{4}{1}\binom{52-4}{13}+\binom{4}{2}\binom{52-2\cdot4}{13}-\binom{4}{3}\binom{52-3\cdot4}{13}+\binom{4}{4}\binom{52-4\cdot4}{13}.$$

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  • $\begingroup$ Oh I read the question wrong. Can you explain how you used inclusion exclusion exactly? $\endgroup$ – Ashuv Halim Feb 9 '18 at 10:43
  • $\begingroup$ Exactly this one $\endgroup$ – Ashuv Halim Feb 9 '18 at 10:46
  • $\begingroup$ There should be more elements using Inclusion-Exclusion, no? How did you get to that calculation in the end? $\endgroup$ – Ashuv Halim Feb 9 '18 at 10:52
  • $\begingroup$ Note that $|J|=|Q|=|K|=|A|=\binom{52-4}{13}$, and $|J\cap Q|=|J\cap K|=|J\cap A|=|Q\cap A|=\dots=\binom{52-2\cdot 4}{13}$, ... $\endgroup$ – Robert Z Feb 9 '18 at 10:54

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