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?
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, $$\binom{52}{13}-\binom{4}{1}\binom{52-4}{13}+\binom{4}{2}\binom{52-2\cdot4}{13}-\binom{4}{3}\binom{52-3\cdot4}{13}+\binom{4}{4}\binom{52-4\cdot4}{13}.$$