I have a deck of 12 cards - one Jack, Queen and King of each suit. There are 5 cards in one hand. How many hands are there in which a Jack, Queen and King all show up and all 4 suits show up? My thinking behind this:
Number of hands = (Number of ways to choose the 3 suits that we get the Jack, Queen and King from) $\times (3!) \times$ (Number of ways to choose any card from the 4th suit) $\times$ (Number of ways of getting any 5th card) = $\binom{4}{3}\times3!\times\binom{3}{1}\times\binom{8}{1} = 576$.
I'm not sure if this is right as the total seems too large. Please confirm/correct where I am going wrong, thanks.