1. How many different triangles can be formed from: a) 40 matches; b) 43 matches? Solution. We need to find the number of triples of natural numbers $\mathrm{x}, \mathrm{y}, \mathrm{z}$ such that $\mathrm{x} \leq \mathrm{y} \leq \mathrm{z}$, $\mathrm{x}+\mathrm{y}+\mathrm{z}=40$ and $\mathrm{x}+\mathrm{y}>\mathrm{z}$. From these inequalities, it follows that $\mathrm{z}$ can take values satisfying the inequalities $14 \leq z \leq 19$. If $z=19$, then $x+y=21$, and $x \leq y \leq 19$. Therefore, $11 \leq y \leq 19$, and we have 9 triangles with $z=19$. Similarly, we establish that the number of triangles for which $\mathrm{z}=18,17,16,15,14$ is respectively $8,6,5,3,2$, and in total we have 33 triangles. Similarly, we find that the number of triangles with a perimeter of 43 is 44.
Answer: a) $33 ;$ b) 44.