How many ways can 32 cards be arranged in 4 rows (with eight cards each)?
I. Solution. We consider two arrangements different if in at least one row, at least one card is different from the others, while the rest are the same. It is natural that in this case, there will also be a difference in at least one other row.

Let $x$ be the number of such arrangements. If we select 8 specific cards for each row and permute them within their own row, 8! permutations will result from each row; each of these can be combined with the permutations of the other rows, so in total, $(8!)^4$ combinations are possible, with the same cards remaining in each row. If we do this for each of the $x$ cases, we will have obtained all permutations of the 32 elements, that is,

$$
x \cdot (8!)^4 = 32! \text{ and hence } x = \frac{32!}{(8!)^4}
$$

II. Solution. From 32 cards, we can choose 8 for the first row in $\binom{32}{8}$ ways. For the second row, we need to choose 8 from the remaining 24; this is possible in $\binom{24}{8}$ ways. For the third row, we need to choose 8 from the remaining 16; this can be done in $\binom{16}{8}$ ways. For the fourth row, there are $\binom{8}{8}$ ways to choose. Thus, the total number of arrangements is

$$
\binom{32}{8}\binom{24}{8}\binom{16}{8}\binom{8}{8} = \frac{32!}{8!24!} \cdot \frac{24!}{8!16!} \cdot \frac{16!}{8!8!} \cdot \frac{8!}{8!} = \frac{32!}{(8!)^4}
$$

István Székely (Berzsenyi Dániel g. VIII. o. Bp. V.)

Note. If we also consider the order within each row, then, as we see from the formulation of the first solution, the number of arrangements is 32!