**Topic**

A football tournament consists of 16 teams, including 4 teams from country V. The organizers draw randomly to divide into 4 groups A, B, C, D, each group has 4 teams. Calculate the probability of the event “Four teams of country V in 4 different groups”

**Solution method – See details**

The probability of event A being a number, symbol \(P\left( A \right)\) is determined by the formula: \(P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}}\), where \(n\left( A \right)\) and \(n\left( \Omega \right)\) denote the number of elements of set A and \(\Omega \) respectively.

**Detailed explanation**

+ Place 16 teams into 4 groups of 4 players each

Choose 4 people from 16 people, then choose 4 people from the remaining 12 people, then choose 4 people from the remaining 8 people.

\( \Rightarrow n\left( \Omega \right) = C_{16}^4.C_{12}^4.C_8^4\)

+ Let A be the event “Four teams of country V in 4 different groups”

+ The number of ways to place 4 teams of country V in the group is \(4!\)

+ Number of ways to arrange the remaining 12 teams into 4 groups: \(C_{12}^3.C_9^3.C_6^3\)

\( \Rightarrow n\left( A \right) = 24.C_{12}^3.C_9^3.C_6^3\)

Then the probability of event A is:

\(P\left( A \right) = \frac{{n\left( A \right)}}{{n\left( \Omega \right)}} = \frac{{24.C_{12}^ 3.C_9^3.C_6^3}}{{C_{16}^4.C_{12}^4.C_8^4}} = \frac{{64}}{{455}}\)