Question

In a tournament, there are $n$ teams, $T_1,T_2,T_3,...,T_n$, with $n>5$. Each team consist of $K$ players $K>3$. The following pairs of teams have one player in common $T_1$ and $T_2,T_2$ and $T_3,...,T_{n-1}$ and $T_n$ and $T_n$ and $T_1$. No other pair of teams has any player in common. How many players are participating in the tournament, considering all the $n$ terms together?

• A. $K(n-1)$
• B. $n(K-2)$
• C. $K(n-2)$
• D. $n(K-1)$

Answer 1

The correct option is B $n(K-2)$

Number of teams $=n(n>5)$
Number of player in $1$ team $=K(K>3)$
Now, consider team $1$
It has number of player $=k(k>3)$
Now, number of player common with the other team $(T_n,T_2)=2$
So, number of uncommon player in each team $=K-2$ and number of teams $=n$
$\therefore$ Total number of players $=n(K-2)$.