Task B-3.2. Let the set $A=\{n: n \in \mathbb{N}, n<101\}$. How many four-element subsets of the set $A$ are there such that the difference between the largest and smallest element is 12?
Solution. We consider the four-element subsets of the given set.

The difference between the largest and smallest element will be 12 in the following cases:

1) If we choose the numbers 1 and 13, the remaining two numbers can be any two numbers between 1 and 13. We can choose them in $\frac{11 \cdot 10}{2}$ ways. We divide by two because, for example, the numbers 2, 8 and 8, 2 are the same choice, and we counted them twice.
2) If we choose the numbers 2 and 14, the remaining two numbers between 2 and 14 can be chosen in $\frac{11 \cdot 10}{2}$ ways.
3) If we choose the numbers 3 and 15, the remaining two numbers between 3 and 15 can be chosen in $\frac{11 \cdot 10}{2}$ ways.
4) If we choose the numbers 88 and 100, the remaining two numbers between 88 and 100 can be chosen in $\frac{11 \cdot 10}{2}$ ways.

The total number of possibilities is

$$
88 \cdot \frac{11 \cdot 10}{2}=4840
$$