5.090 In a chess match between two teams of 8 people, the participants of the games and the color of the pieces for each participant are determined by a draw. What is the number of different outcomes of the draw?
Solution. For each pair of players, there are 2 ways to distribute the color of the pieces. There are a total of 8 pairs. Therefore, the total number of ways to distribute the color of the pieces is $2^{8}$.

Number the players of the first team as $1, 2, \ldots, 8$. Then player №1 can be paired with one of 8 players from the other team, player №2 with one of the 7 remaining players from the other team, and so on, until player №8 is paired with the last remaining player from the other team. In total, we get $8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 8!$ possible ways to form pairs. Therefore, the total number of possible outcomes of the draw is $2^{8} \cdot 8!$.

Answer: $2^{8} \cdot 8!$

## REFERENCE MATERIALS

This reference material contains a multitude of formulas and properties selected in accordance with the main methodological principle of this book. The essence of this principle lies in the strict adherence to the equivalence of all transformations performed when solving equations and inequalities of any type. This principle allows for uniformity and logic in solving tasks of each section, without being distracted by labor-intensive work with cumbersome domains of definition and other, often excessive, restrictions on variables.

However, in many traditional formulas, the domains of definition of the left and right parts do not coincide. Therefore, such formulas are presented here in the form of transformations that preserve equivalence (within the corresponding formula). In this case, identical transitions are marked with the symbol $\Leftrightarrow$.

## I. Formulas for Simplified Multiplication

$$
\begin{aligned}
& (a+b)^{2}=a^{2}+2 a b+b^{2} \\
& (a-b)^{2}=a^{2}-2 a b+b^{2} \\
& (a+b)(a-b)=a^{2}-b^{2} \\
& (a+b)\left(a^{2}-a b+b^{2}\right)=a^{3}+b^{3} \\
& (a-b)\left(a^{2}+a b+b^{2}\right)=a^{3}-b^{3} \\
& (a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)=a^{3}+b^{3}+3 a^{2} b+3 a b^{2} \\
& (a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)=a^{3}-b^{3}-3 a^{2} b+3 a b^{2}
\end{aligned}
$$

## Vieta's Theorems