4.1. A teacher at a summer math camp took with him for the whole summer several shirts, several pairs of pants, several pairs of shoes, and two jackets. At each lesson, he wore pants, a shirt, and shoes, and he wore a jacket on some lessons. On any two lessons, at least one of the items of his clothing or shoes was different. It is known that if he had taken one more shirt, he could have conducted 18 more lessons; if he had taken one more pair of pants, he could have conducted 63 more lessons; if he had taken one more pair of shoes, he could have conducted 42 more lessons. What is the maximum number of lessons he could conduct under these conditions?
Answer: 126.

Solution: Let the teacher bring $x$ shirts, $y$ pairs of trousers, $z$ pairs of shoes, and 2 jackets. Then he can conduct $3 x y z$ lessons (the number 3 means: 2 lessons in each of the jackets and 1 lesson without a jacket). If he has one more shirt, the number of lessons will increase by $3 y z$. If he has one more pair of trousers, the number of lessons will increase by $3 x z$. If he has one more pair of shoes, the number of lessons will increase by $3 y z$. Thus, we get a system of three equations: $3 y z=18, 3 x z=63, 3 x y=42$. Therefore, $y z=6, x z=21, x y=14$, and thus $(x y z)^{2}=6 \cdot 21 \cdot 14, x y z=42$ (although it is not necessary, but we can calculate that $x=7, y=2, z=3$). The desired value: $3 x y z=126$.

Remark. There is a possible interpretation of the problem condition in which it is required to find the maximum number of lessons given that the teacher will take with him one more shirt, one more pair of trousers, and one more pair of shoes. The corresponding answer $3(x+1)(y+1)(z+1)=288$ is also considered correct.