## Solution 36 Page 48 Math Workbook 10 – Kite>

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:

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