12 different items are distributed among 3 people so that each person gets 4 items. In how many ways is this possible?
I. solution. The first item that goes to the first person $E$ can be chosen in 12 ways. Regardless of which item was chosen first, the second item can be chosen from the remaining ones in 11 ways. Similarly, the third and fourth items can be chosen in 10 and 9 ways, respectively, for $E$. This gives $N_{1}=12 \cdot 11 \cdot 10 \cdot 9$ possibilities for the distribution of these 4 items. This includes the order in which the items are given.

However, the order is irrelevant; $E$ will receive the same items regardless of whether they are given in the order $A, B, C, D$ or $B, D, C, A$, or any other order. Similar reasoning shows that 4 different items can be given in $4 \cdot 3 \cdot 2=24$ different orders, i.e., in this many ways we can choose which to give first, second, and third, with the last one being determined. Therefore, the number of distinct ways to satisfy $E$ is

$$
N_{1}=\frac{12 \cdot 11 \cdot 10 \cdot 9}{4 \cdot 3 \cdot 2}=11 \cdot 5 \cdot 9=495
$$

Similarly, the second person $M$ can receive their share from the remaining 8 items in

$$
N_{2}=\frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2}=70
$$

different ways. This completes the distribution, with the third person $H$ receiving the remaining items. Thus, the distribution can be done in $N_{1} \cdot N_{2}=34650$ ways.

László Szilágyi (Debrecen, Fazekas M. g. IV. o. t.)

II. solution. Prepare 4 cards each with the letters $E, M$, and $H$ to mark the items to be given to the 3 people. For now, distinguish the cards with the same letter by indexing them as $1,2,3,4$: $E_{1}, E_{2}, E_{3}, E_{4}, M_{1}, \ldots, H_{4}$.

Similar reasoning to the first solution shows that the 12 cards can be placed on the (arranged) 12 items in $N_{3}=12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2$ ways. Each placement represents a distribution possibility, but not all are distinct. The 4 $E$-lettered cards can be swapped among the 4 items without changing $E$'s share, and as seen in the first solution, there are $N_{4}=4 \cdot 3 \cdot 2$ ways to do this. Repeating this reasoning for $M$ and $H$, we find that each distinct distribution appears $\left(N_{4}\right)^{3}$ times in the $N_{3}$ number of distributions. Thus, the number of distinct distributions is:

$$
\frac{N_{3}}{\left(N_{4}\right)^{3}}=\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}{4 \cdot 3 \cdot 2 \cdot 4 \cdot 3 \cdot 2 \cdot 4 \cdot 3 \cdot 2}
$$

After simplifying the last three factors in the numerator and the denominator, we see that our expression is equal to the product $N_{1} N_{2}$ from the first solution.

Angela Nagy (Balassagyarmat, Szántó Kovács J. g. IV. o.t.)

Note: Many contestants provided answers using combinatorial formulas, despite the fact that the rules of the competition - K.M.L. 25 (1962) p. 15 - do not accept mere references to theorems not covered in the high school curriculum, as the thought process is more important than the ready-made template. Nevertheless, we accepted as solutions those papers that justified why the formulas they used led to the correct result. (We also point out: there is a large number of incorrect papers, in which the thought process is missing or incorrect.)