A 25-meter long coiled cable is cut into pieces of 2 and 3 meters. In how many ways can this be done if the order of the different sized pieces also matters?
Let the number of 2-meter pieces be $x$, and the number of 3-meter pieces be $y$. First, we need to solve the following Diophantine equation:

$$
2 x+3 y=25
$$

From the equation,

$$
x=\frac{25-3 y}{2}
$$

Since we are looking for integer solutions, $25-3 y$ must be even. This means that $3 y$ and thus $y$ must be odd. Furthermore, since $x>0$, $3 y<25$. Therefore, $y$ must be a positive odd number less than 8; its possible values are: $1, 3, 5, 7$.

The corresponding values of $x$ are: $\quad 11, 8, 5, 2$.

In the first case, we have a total of 12 pieces of wire (1 piece of 3 meters and 11 pieces of 2 meters), and the number of possible arrangements is $\frac{12!}{1!\cdot 11!}=12$.

For the other cases, similarly: $\frac{11!}{3!\cdot 8!}=165$, $\frac{10!}{5!\cdot 5!}=252$, and $\frac{9!}{2!\cdot 7!}=36$.

Thus, the total number of possible cases is: $165+252+36+12=465$.

Krisztian Balog (Miskolc, Foldes F. Gymnasium, 12th grade)