# Task 9. Combinatorics

Problem text. 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, 13 oil refineries (NPR) and 15 gas transportation hubs (GTU) are selected. 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.

A total of 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 we exclude them from the total number) 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 55 ways. From 12 processing units, we can select three in 220 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 = **12,100** ways.

Problems in other variants are similar, so the solution is provided for only one variant.