Question

In a football tournament, a team $T$ has to play with each of the $6$ other teams once. Each match can result in a win, draw or loss. The number of ways in which the team $T$ finishes with more wins than losses, is

• A. $392$
• B. $294$
• C. $129$
• D. $213$

Answer 1

The correct option is B $294$

Total number of results $=3^6=729$

Required number of ways $=\frac{1}{2}(729-\left(\begin{array}{c}\text{Number of ways in which team}T\text{finishes}\\ \text{with equal number of wins and losses}\end{array}\right))$

Now, we shall consider following cases :

Case $\text{I}:$ $0$ draw, $3$ wins and $3$ losses
$WWWLLL$
Number of ways $=\frac{6!}{3!3!}={}^6{C}_3=20$

Case $\text{II}:$ $1$ win, $1$ loss, $4$ draws
$WLDDDD$
Number of ways $=\frac{6!}{4!}=30$

Case $\text{III}:$ $2$ wins, $2$ losses, $2$ draws:
$WWLLDD$
Number of ways $=\frac{6!}{2!2!2!}=90$

Case $\text{IV}:$ no win and no loss, $6$ draws :
$DDDDDD$
Number of ways $=1$

Total ways with equal wins and draws $=20+30+90+1=141$
So, required number of ways $=\frac{1}{2}(729-141)=\frac{588}{2}=294$