Anikó, Béla, Csaba, and Dóra took out the figures from a deck of French cards. They distribute the remaining cards, each getting four. Then the bank pays $1 \mathrm{Ft}$ for each jack, $3 \mathrm{Ft}$ for each queen, $5 \mathrm{Ft}$ for each king, and $7 \mathrm{Ft}$ for each ace. In how many ways can it happen that everyone gets 16 Ft?
Let's denote the ace by $A$, the king by $K$, the queen by $D$, and the jack by $B$. It is easy to see that the value of the four-card set is 16 in the following five cases:

$$
\begin{array}{lllll}
\text { a) } A K D B, & \text { b) } A D D D, & \text { c) } K K K B, & \text { d) } K K D D, & \text { e) } A A B B \text {. }
\end{array}
$$

Next, we look for those four-card sets within which there are a total of $4 A, 4 K, 4 D, 4 B$. I. The c) four-card set can only appear once and cannot be followed by d), because then there would be 5 kings, but a) must be there—due to the 4 kings—and to have four of each figure, there must also be a b) type and an e) type four-card set. In any other distribution, the c) type four-card set cannot appear.

If we start from the b) case, by examining the number of queens, we get the same result. Therefore, the b) type cannot appear in more distributions. II. From the d) type four-card set, we can take $2 \text{db}-\text{of}$, and only $2 e$ can stand next to them, because e) contains neither K nor D.

If there is only one d) type four-card set, then to have $4 K$ and $4 D$ in the distribution, 2 of the remaining 3 players must receive a) type four-card sets, and only 2. For a good distribution, the fourth can only receive e).

If we start the II. examination with the distribution of the e) type four-card set and examine the number of aces and jacks, we arrive at the same result. Neither the d) nor the e) type four-card set can appear in more distributions.

Finally, if we start with the a) type four-card set and distribute four of them, we get a good distribution. Thus, the four possible distributions are:

$$
\text { 1. cabe, } \quad \text { 2. ddee, } \quad \text { 3. daae, } \quad \text { 4. aaaa. }
$$

Naturally, a different distribution is counted if the four cards are exchanged among the four players. The number of possible such distributions in the 1st case is $4!=24$, in the 2nd case $\frac{4!}{2!2!}=6$, in the 3rd case $\frac{4!}{2!}=12$, and in the 4th case $\frac{4!}{4!}=1$, so in total 43 good distributions exist.

If we also distinguish the suits of the individual cards, then the colors can be combined in the following way for a given figure distribution:

In the 1st case, the first player can receive cards in $4^{4}$ ways, the second in $\binom{3}{2}\binom{3}{2}$ ways, the third and fourth in one way. This results in 2304 cases in total.

In the 2nd case, the first player can receive cards in $\binom{4}{2}\binom{4}{2}$ ways, the second in $\binom{2}{2}\binom{2}{2}$ ways, the third in $\binom{4}{2} \binom{4}{2}$ ways, and the fourth player's cards are given. This results in 1296 cases in total.

In the 3rd case, similarly, $4^{4} \cdot 3^{4}=20736$ different distributions exist, and in the 4th case, $4^{4} \cdot 3^{4} \cdot 2^{4} \cdot 1^{4}=331776$.

Including the exchange of cards among the players, in total $24 \cdot 2304 + 6 \cdot 1296 + 12 \cdot 20736 + 1 \cdot 331776 = 643680$ different distributions satisfy the requirements described in the problem. (V. A.)