Péter, Rezső, Sándor, and Tamás together have 6 drawing triangles. In how many ways is this possible? (It is not certain that each of them brought a drawing triangle, for example, P-, R-, S-, T- could have $2,0,3,1$ respectively.)
I. solution. Without regard to the order (i.e., the names), the distribution of 6 drawing triangles among 4 people can be as follows:
a) $6,0,0,0$
b) $5,1,0,0$
c) $4,2,0,0$
d) $4,1,1,0$
e) $3,3,0,0$
f) $3,2,1,0$
f) $2,2,1,1$.
g) $3,1,1,1$
h) $2,2,2,0$;

they are distributed equally. Or like this: someone has $x$ and the others have $y$ number of drawing triangles, - where of course $x+3y=6$, and $x \neq y$. Let's call the sequence $x, y, y, y$ the formula of the distribution. The "someone" (the $x$) in each of the three cases can be any of the 4 people, so $3 \cdot 4=12$ such distributions exist.

$\beta)$ The cases b), c), and d) are the same in that 2 people have the same number of drawing triangles, and the other 2 have a different number of drawing triangles from each other and from the first 2, the distribution formula is: $x, y, z, z$, where $x \neq y \neq z \neq x$. The first person, who has a different ( $x$ ) number of drawing triangles, can be chosen in 4 ways. Keeping this person unchanged, the second (the $y$) can be chosen from the remaining in 3 ways. This already determines who the $z$ number of drawing triangles belong to, so the people can be assigned to the distribution in $4 \cdot 3=12$ ways in each of the three cases. There are $3 \cdot 12=36$ such formula distributions.

ү) The cases e) and j) differ from the previous ones only in that $x=y$. Therefore, 2 of the previous possibilities are the same. For example, if Peter had $x$ and Rezso had $y$ before, now the "Rezso, Peter" order gives the same distribution. Thus, in both cases, there are $12:2=6$ possibilities, together $2 \cdot 6=12$.

ס) Finally, the f) distribution is unique, it is only possible for everyone to have a different number of drawing triangles. Selecting the owners of the 3, 2, and 1 drawing triangles in sequence, the first can be chosen in 4 ways, the second in 3 ways, and the third in 2 ways, so the number of such distributions is $4 \cdot 3 \cdot 2=24$.

Thus, the number of distributions from the $\alpha$), $\beta$), $\gamma$), $\delta$) groups is $12+36+12+24=84$.

András Hirka (Pannonhalma, Bencés g. I. o. t.)

II. solution. If there were only two boys, regardless of how many drawing triangles they had together, this could always be distributed in one more way than the number of drawing triangles. Indeed, one boy could have all the drawing triangles, or one less, or two less, and so on, or 2; or 1, and finally, none at all. From this, we always see the number of drawing triangles the other boy has.

Similarly, in our problem, first dividing the four boys into two pairs, the 6 drawing triangles can be distributed in 7 ways between the pairs, the first pair can have

$$
6,5,4,3,2,1, \quad \text { or } \quad 0
$$

drawing triangles, and the remaining

$$
0,1,2,3,4,5,6
$$

drawing triangles belong to the second pair.

Now suppose Peter and Rezso together have 4 drawing triangles, so the other two together have 2 left. In the first pair, there are 5, and in the second pair, there are 3 possible distributions, and since each distribution in the first pair can be combined with each distribution in the second pair, there are $5 \cdot 3=15$ possible distributions among the 4 boys under our assumption.

Going through all the paired possibilities for the 4 boys in (1) and (2)

$$
7 \cdot 1+6 \cdot 2+5 \cdot 3+4 \cdot 4+3 \cdot 5+2 \cdot 6+1 \cdot 7=84
$$

distribution possibilities are obtained.

András Szép (Budapest Rákóczi F. g. I. o. t.)

III. solution. Let's visualize the distribution possibilities as follows. Draw a segment of unit length (horizontal) next to each boy's name, the length of which is the number of drawing triangles he has, and if he has no drawing triangles, draw a segment of zero length (i.e., a point). Place these segments one below the other, at a unit distance from each other, and then move the starting point of each segment to the point directly below the end point of the previous one, starting with Rezso.

![](https://cdn.mathpix.com/cropped/2024_05_02_b23801cfd24f53d25611g-1.jpg?height=304&width=550&top_left_y=2339&top_left_x=771)

Figure 1

Thus, the $V$ endpoint of Tamás's segment is 6 units to the right and 3 units down from the $K$ starting point of Péter's segment. Drawing our diagrams on grid paper and connecting the end and starting points of adjacent segments with lines that always go either to the right or down, we get a path from $K$ to $V$ that always follows the grid lines. The distribution example mentioned in the problem text is illustrated by Figure 1.

A different distribution corresponds to a different such path, and conversely, every path from $K$ to $V$ that follows the perimeter or internal grid lines of the rectangle $KLVU$ in Figure 1 corresponds to a distribution.

Thus, we only need to determine how many ways there are to get from $K$ to $V$ while maintaining the given conditions. This can be obtained by writing the number of paths leading to each node of the grid in Figure 2.

![](https://cdn.mathpix.com/cropped/2024_05_02_b23801cfd24f53d25611g-2.jpg?height=312&width=468&top_left_y=516&top_left_x=796)

Figure 2

The first row and the first column of every node can be reached in only one way without changing direction. Every other node can be reached either from the node above it or from the node to its left, so the number of ways to reach it is equal to the sum of the numbers written to those nodes. Thus, there are 84 paths to $V$, and this many distributions are possible.