Question

There are 16 teams divided in 4 groups. every team from each group will play with each other once. the top 2 teams will go to the next round and so on the top two teams will play the final match. minimum how many matches will be played in that tournament?

Answer 1

aww.... seems question is based on permutation and combination. Let's make it easy to solve.


according to question,
there are 16 team are divided into 4 groups , means each group has 4 teams.every team from each group will play with each other once.

Let first group contains P, Q, R ,S team.
then they can play with them self = $^4C_2$ ways
= 4!/2!(4 - 2)! = 4 × 3 × 2!/2 × 2! = 6 ways

similarly, 2nd group, team can play with them self = 6 ways
3rd group, team can play with them self = 6 ways
4th group, team can play with them self = 6 ways

now, from every group , top 2 team is selected.
so, total 8 team are selected again they form two group , each group has 4 team.
they can again arrange .
first group, team can play with them self = 6 ways
2nd group, team can play with them self = 6 ways


form this two group again selected top two team , total 4 team again arrange in 6 different ways.
from this two team selected and these two team play final match e.g., 1 way

now total number of ways = (6 + 6 + 6 + 6) + (6 + 6) + 6 + 1 = 24 + 12 + 7 = 36 + 7 = 43

hence, answer should 43

Answer 2

Answer:
16 Teams

4 Groups
Each group 4 teams

so Each single group will play with each other
so that is 6 matches per group that is 6*4= 24 matches
then 8 teams are eliminated from the sixteen the enter in  quarters finals
where they will play each other so 8/2=4 matches

then four other teams are eliminated this round
now four teams are remaining
4/2= 2 matches
where 2 teams are out in semi finals
2/2=1 match

Now = 24+4+2+1=
31 matches the final answer

Step-by-step explanation: