10 tennis players arrived at the competition, 4 of whom are from Russia. According to the rules, for the first round, the tennis players are randomly paired. Find the probability that in the first round all Russian players will be paired only with other Russians.
Let's imagine that instead of pairing up for a game, we want to accommodate 10 tennis players in pairs in five double rooms in a hotel, which is the same from a probability theory perspective. The question of the problem boils down to the probability that four Russian athletes will occupy exactly two rooms.

Schematically, we can represent the double rooms with five rectangles. Each rectangle is divided in half (since there are two athletes in a room). It doesn't matter which specific Russian athlete occupies which place. What matters is that she is Russian. Therefore, the places occupied by Russians can be marked identically. For example, with a flag. We will not mark the other places.
![](https://cdn.mathpix.com/cropped/2024_05_08_22585010b2534c7ab9ebg-43.jpg?height=146&width=954&top_left_y=2456&top_left_x=550)

In the image, one of the arbitrary accommodations is shown - some four places out of the ten available have been chosen. There are $C_{10}^{4}$ ways to do this.

Now let's consider the accommodation where the four Russians occupy exactly two rooms. Such a situation is shown in the following image. There are exactly $C_{5}^{2}$ such accommodations, as the enumeration has been reduced to choosing two rooms out of five.

Therefore, the probability of getting two "purely Russian pairs" is $\frac{C_{5}^{2}}{C_{10}^{4}}=\frac{5!\cdot 4!\cdot 6!}{2!3!\cdot 10!}=\frac{4 \cdot 3 \cdot 6 \cdot 5 \cdot 4}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}=\frac{1}{21} \approx 0.048$.

## Answer

$1 / 21$

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