Hello, chopet!

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.

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