How many bridge hands have a 5-card suit that must contain the ace of that suit, a 4-card suit, and a void (no cards of a suit) How many bridge hands have a 5-card suit that must contain the ace of that suit, a 4-card suit, and a void (no cards of a suit)? I only know the first two are: C(12,4)C(13,4), but I really don't know what the "void" means and how to deal with it in the question. Thanks!

The distribution is of the type $5$-$4$-$4$-$0$. (We are **void** in a suit, say spades, if we have $0$ spades.)

The suit we have $5$ of can be chosen in $\binom{4}{1}$ ways. For each such choice, as you saw, the actual cards can be chosen in $\binom{12}{4}$ ways, since one of the cards must be an Ace. For each such choice, the suit we are void in can be chosen in $\binom{3}{1}$ ways. And then the cards in the two suits we have $4$ of can be chosen in $\binom{13}{4}^2$ ways. Multiply.