Question

Three players play a total of 9 games. In each game, one person wins and the other two lose; the winner gets two 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

Hint: It is given that three players play a total of 9 games and in each game, one person wins and the other two lose. The points division is done in such a way that the winner gets two points and the loser loses 1 point. Each player need to win three matches and lose 6 matches to get a zero score so the number of ways in which this can be true is by selecting 3 matches out of 9 matches using combinatorial approach then selecting 3 matches out of 6 matches using combinatorial approach and finally selecting 3 matches out of 3 matches using combinatorial approach. After that, multiply all these selections to one another.

Complete step-by-step answer:
There are 9 games to be played by 3 players. If a person wins then that person is awarded with 2 points and if a person loses then that person will lose 1 point. We have to find the number of ways in which all the 9 games are played such that each of the players will finish it with a zero.
The score of each player will be 0 when each player wins 3 matches and loses 6 matches. To achieve that, we are going to select three matches which are going to be won by first player from 9 matches using combinatorial method:
${}^{9}{{C}_{3}}$ ………… (1)
Now, selecting 3 matches out of 6 matches by second player using combinatorial approach we get:
${}^{6}{{C}_{3}}$ ……….. (2)
Selecting 3 matches out of 3 matches by third player using combinatorial approach we get:
${}^{3}{{C}_{3}}$ …………… (3)
Multiplying (1), (2), (3) we get:
${}^{9}{{C}_{3}}\left( {}^{6}{{C}_{3}} \right){}^{3}{{C}_{3}}$
We know that:
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Using the above relation in simplifying the combinatorial terms we get,
$\begin{align}
  & \dfrac{9!}{3!\left( 9-3 \right)!}\left( \dfrac{6!}{3!\left( 6-3 \right)!} \right)\left( \dfrac{3!}{3!\left( 3-3 \right)!} \right) \\
 & =\dfrac{9!}{3!6!}\left( \dfrac{6!}{3!3!} \right)\left( \dfrac{3!}{3!0!} \right) \\
 & =\dfrac{9.8.7.6!}{3!6!}\left( \dfrac{6.5.4.3!}{3!3!} \right)\left( \dfrac{3!}{3!\left( 1 \right)} \right) \\
\end{align}$
In the above expression, $6!\And 3!$ will be cancelled out from the numerator and the denominator and we get,
$\begin{align}
  & \dfrac{9.8.7}{3!}\left( \dfrac{6.5.4}{3!} \right)\left( \dfrac{1}{\left( 1 \right)} \right) \\
 & =\dfrac{9.8.7}{3.2.1}\left( \dfrac{6.5.4}{3.2.1} \right) \\
 & =1680 \\
\end{align}$
Hence, the number of ways in which they can play all the 9 games and finish each with a zero score is 1680 and the correct option is (b).

So, the correct answer is “Option (b)”.

Note: In the above, we have solved ${}^{9}{{C}_{3}}\left( {}^{6}{{C}_{3}} \right){}^{3}{{C}_{3}}$ by using the following relation:
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Instead of using the above relation, we can solve ${}^{9}{{C}_{3}}\left( {}^{6}{{C}_{3}} \right){}^{3}{{C}_{3}}$ in the following way:
While solving ${}^{9}{{C}_{3}}$ we can write in the numerator 9.8.7 and in the denominator just $3!$.
$\dfrac{9.8.7}{3!}$
Similarly, solving ${}^{6}{{C}_{3}}$ we can write numerator as 6.5.4 and denominator as $3!$ we get,
$\dfrac{6.5.4}{3!}$
And writing ${}^{3}{{C}_{3}}$ as 1 because the value of ${}^{n}{{C}_{n}}$ is 1.
Now, multiplying the above three we get,
$\begin{align}
  & \dfrac{9.8.7}{3!}\left( \dfrac{6.5.4}{3!} \right)1 \\
 & =\dfrac{9.8.7}{6}\left( \dfrac{6.5.4}{6} \right) \\
 & =1680 \\
\end{align}$
As you can see that we are getting the same answer which we are getting in the above solution.