For each such company, associate a set of 10 people, which includes all the girls who joined the company and all the boys who did not join it.

## Solution

Let there be some company consisting of $k$ boys and $k$ girls. We will associate it with a set of 10 people, which includes $k$ girls who joined the company and $10-k$ boys who did not join it. The established correspondence is clearly one-to-one. Therefore, the desired number is equal to the number of ways to choose 10 people out of 20.

## Answer

$C_{20}^{10}=184756$ ways.
$\left[\begin{array}{l}\text { Combinations and Permutations } \\ {\left[\begin{array}{l}\text { Partitions into Pairs and Groups; Bijections }]\end{array}\right]}\end{array}\right]$

There are 20 people - 10 boys and 10 girls. How many ways are there to form a company in which there would be an equal number of boys and girls?