Question

In the next world cup of cricket there will be 12 teams, divided equally in two groups. Teams of each group will play a match against each other. From each group 3 top teams will qualify for the next round. In this round each team will play against others once. Four top teams of this round will qualify for the semifinal round, when each team will play against the others once. Two top teams of this round will go to the final round, where they will play the best of three matches. The minimum number of matches in the next world cup will be
(a) 54
(b) 53
(c) 52
(d) None of these

Answer 1

Hint: There are 12 teams divided into two groups. So, we get the number of matches in the first round by using a combination formula. From each group, three top teams will qualify for the next round and each team will play against another one. Hence, we get a number of matches in the next round. Top 4 from this round are selected and one team will play against the other three. We also get a number of matches in the semifinal round. At last, two top teams of this round will go to the final round, where they will play the best of three matches. Adding all the rounds matches which we found earlier, we get the total number of matches.

Complete step-by-step answer:
Total number of teams which play the world cup = 12 teams
One important formula used in solving is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
All 12 teams are divided in two groups. In one group we have 6 teams and 2 teams will play one match. So, for two groups having 6 teams. Putting n = 6 and r = 2, we get
Matches in the first round:
 $\begin{align}
  & ={}^{6}{{C}_{2}}+{}^{6}{{C}_{2}} \\
 & =\dfrac{6!}{4!2!}+\dfrac{6!}{4!2!} \\
 & =\dfrac{6\times 5\times 4!}{4!2!}+\dfrac{6\times 5\times 4!}{4!2!} \\
 & =15+15 \\
 & =30 \\
\end{align}$
From each group three top teams will qualify for the next round and each team will play against another one. So, we have only 6 teams left after the first round and 2 teams will play a match. By using combination formula and putting n = 6 and r = 2, we get
Match in the second round
$\begin{align}
  & ={}^{6}{{C}_{2}} \\
 & =\dfrac{6!}{4!2!} \\
 & =\dfrac{6\times 5\times 4!}{4!2!} \\
 & =15 \\
\end{align}$
We get top 4 from this round where each team will play against the other three. Now, putting n = 4 and r = 2 in combination formula, we get
Matches in the semi finals
$\begin{align}
  & ={}^{4}{{C}_{2}} \\
 & =\dfrac{4!}{2!2!} \\
 & =\dfrac{4\times 3\times 2!}{2!2!} \\
 & =6 \\
\end{align}$
Final is best of three so if one team wins the first match and second match then there will not be a third match. Hence, the minimum number of matches are:
The total number of matches = Matches in the first round + Matches in the second round + Matches in the semi finals round + 2 matches in finals
Total number of matches $=30+15+6+2=53$
Hence, the number of minimum matches in next year's world cup is 53.
Hence, option (b) is correct.

Note: It is mentioned that we are required to find the minimum number of matches. So, one common mistake is the consideration of all three matches in the final. Since it is given that the final matches are best of three. By using this, we have to predict the correct number of matches in the final. Knowledge of combination is must for solving this problem.