In how many ways can 10 and 50 fillér coins be placed side by side (all centers on a straight line) to cover a $1 \mathrm{~m}$ long segment (not more), if at least 50 coins are used, and the order of the two types of coins is taken into account. (Coins of the same value are not distinguished. The diameter of the 10 fillér is $19 \mathrm{~mm}$, and that of the 50 fillér is $22 \mathrm{~mm$).
Suppose that to cover a $1 \mathrm{~m}=1000 \mathrm{~mm}$ segment, we use $x$ pieces of 10 filléres and $y$ pieces of 50 filléres.

Thus, the conditions

$$
19 x+22 y=1000
$$

and

$$
x+y \geq 50
$$

must be satisfied with positive integers.

From (1),

$$
x=\frac{1000-22 y}{19}=52-y+\frac{12-3 y}{19}=52-y+3 t
$$

where $19 t=4-y$, or

$$
y=4-19 t
$$

and thus

$$
\begin{gathered}
x=52-4+19 t+3 t=48+22 t \\
x+y=52+3 t
\end{gathered}
$$

So,

$$
\begin{gathered}
48+22 t>0 \\
4-19 t>0 \\
52+3 t \geq 50
\end{gathered}
$$

From (3), $t>-\frac{24}{11}$, from (4), $t<\frac{2}{3}$.

All three inequalities are satisfied only by the value $t=0$, so

$$
x=48, \quad y=4 .
$$

52 coins - of which 48 and 4 are equal - are needed to cover the segment. Each covering is a permutation with repetition of these elements, so the number of all possible coverings is

$$
P_{52}^{(48,4)}=\frac{52!}{48!4!}=\frac{52 \cdot 51 \cdot 50 \cdot 49}{1 \cdot 2 \cdot 3 \cdot 4}=13 \cdot 17 \cdot 25 \cdot 49=270725
$$

We can also reason that the position of the 4 (or 48) elements is a 4-class (or 48-class) combination of 52 elements. Generally,

$$
P_{n}^{(k, n-k)}=C_{n}^{k}=C_{n}^{n-k}
$$

István Ujjady (Kecskemét, Piarist High School I. o. t.)