# Task 8. Combinatorics

Text of the problem. At the end of the financial year, a traditional audit of the activities of several subsidiaries of an oil and gas holding is conducted. This year, they include 13 oil refineries (NPR) and 15 gas transportation hubs (GTU). How many options for selecting an audit object exist at the beginning of the second week, if during one week the auditing firm can visit no more than 2 NPRs and 3 GTUs?
Solution to the problem. In total, 13 refineries and 15 processing units need to be inspected. After the first week, 11 refineries remain (since $13-2$ = 11, i.e., 2 refineries were inspected in the first week and were excluded) and 12 processing units ($15-3=12$). To solve this, we can use the combination formula:

$$
\begin{gathered}
C_{11}^{2}=\frac{11!}{2!*(11-2)!}=\frac{11 * 10 * 9!}{2 * 9!}=55 \\
C_{12}^{3}=\frac{12!}{3!*(12-3)!}=\frac{12 * 11 * 10 * 9!}{6 * 9!}=220
\end{gathered}
$$

In other words, from 11 refineries, we can select two in fifty-five ways. From 12 processing units, we can select three in two hundred and twenty ways. Each of the 11 refineries can be combined with each of the 12 processing units. Multiplying these numbers gives the number of ways to select objects for inspection in the second week. 55 * $220=\mathbf{1 2 1 0 0}$ ways.