We need some serious clarification . . .
You said "bridge hands", so I assume 13 cards are dealt to each of four players.
Next, are the four players distinguishable? .Can we call them A, B, C, and D?
If so, there are: . possible ways to deal the cards.
And the problem is far more complicated than you think . . .
Four Aces in one player's hand
There are 4 choices of players to get the four Aces.
There is 1 way for him get the four Aces.
For his other 9 cards, there are: choices.
Then the remaining cards are given to the other three players in: . ways.
Hence, there are: . ways for one player to get the four Aces.
Three Aces in one player's hand, one Ace in another's
There are 4 choices of players to get the three Aces.
There are: ways for him to get three Aces.
For his other 10 cards, there are: ways.
There are 3 choices of players to get the one Ace.
There is one way for him to get the one remaining Ace.
For his other 12 cards, there are: ways.
The remaining 26 cards are distributed to the other two players.
There are: ways.
Hence, there are: . ways
Get the idea?
Edit: corrected an error . . . sorry for the blunder.