Three players play a total of 9 games. In eac | vedclass

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(B) Let the number of games won by the three players be $x, y,$ and $z$ respectively.We are given that the total number of games is $x + y + z = 9$.Th

Vedclass · Accepted Answer

$1680$

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(B) Let the number of games won by the three players be $x, y,$ and $z$ respectively.We are given that the total number of games is $x + y + z = 9$.The total score for each player must be zero at the end.For a player who wins $x$ games and loses the remaining $(9-x)$ games,the total score is $2x - 1(9-x) = 3x - 9$.Setting the score to zero,we get $3x - 9 = 0$,which implies $x = 3$.Similarly,$y = 3$ and $z = 3$.Thus,each player must win exactly $3$ games out of $9$.The number of ways to distribute the wins among the three players is given by the multinomial coefficient:$\frac{9!}{3!3!3!} = \frac{362880}{6 	imes 6 	imes 6} = \frac{362880}{216} = 1680$.

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 get $-1$ each. The number of ways in which they can play all the $9$ games and finish each with a zero score is

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