Question

Three players play a total of $9$ games. In each game, one person wins and the other two lose; the winner gets $2$ points and the losers lose $1$ each. The number of ways in which they can play all the $9$ games and finish each with a zero score is

• A. $84$
• B. $1680$
• C. $7056$
• D. $0$

Answer 1

Answer: (B)

$1680$

Each player need to win $3$ matches and lose $6$ matches to get a zero score.

Out of $9$ games $3$ should be selected by first player(games which he will win)${=}^9{C}_3$

Out of $6$ games 3 should be selected by second player $={}^6{C}_3$

Out of $3$ games $3$ should be selected by third player $={}^3{C}_3$

Hence the answer is $={}^9{C}_3{\times }^6{C}_3{\times }^3{C}_3=1680$