Question

There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is:
(a) 9
(b) 11
(c) 12
(d) 7

Answer 1

Hint: First, before proceeding for this, we must know that for the match to occur between any number of people let us say n people is given by ${}^{n}{{C}_{2}}$. Then, we get the number of matches played between m men in the tournament is given by $2\times {}^{m}{{C}_{2}}$ and we can get the number of matches played between men and women when m men and 2 women are playing in the tournament is given by $2\times {}^{m}{{C}_{1}}\times {}^{2}{{C}_{1}}$. Then, by using the condition given in the question that the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, we get the value of m.

Complete step-by-step solution:
In this question, we are supposed to find the value of m when there are m men and two women participating in a chess tournament and each participant plays two games with every other participant and the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84.
So, before proceeding for this, we must know that for the match to occur between any number of people let us say n people is given by:
${}^{n}{{C}_{2}}$
Then, we get the number of matches played between m men in the tournament is given by:
$2\times {}^{m}{{C}_{2}}$
Similarly, we can get the number of matches played between men and women when m men and 2 women are playing in the tournament is given by:
$2\times {}^{m}{{C}_{1}}\times {}^{2}{{C}_{1}}$
Now, by using the condition given in the question that the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84.
So, by using this condition, we get the expression as:
$2\times {}^{m}{{C}_{2}}-2\times {}^{m}{{C}_{1}}\times {}^{2}{{C}_{1}}=84$
Now, by solving the above expression, we get:
\[\begin{align}
  & 2\times \dfrac{m!}{\left( m-2 \right)!2!}-2\times \dfrac{m!}{\left( m-1 \right)!1!}\times \dfrac{2!}{\left( 2-1 \right)!1!}=84 \\
 & \Rightarrow 2\times \dfrac{m\times \left( m-1 \right)\times \left( m-2 \right)!}{\left( m-2 \right)!2!}-2\times \dfrac{m\times \left( m-1 \right)!}{\left( m-1 \right)!1!}\times \dfrac{2\times 1!}{1!1!}=84 \\
 & \Rightarrow 2\times \dfrac{m\times \left( m-1 \right)}{2}-2\times \dfrac{m}{1}\times \dfrac{2}{1}=84 \\
 & \Rightarrow {{m}^{2}}-m-4m=84 \\
 & \Rightarrow {{m}^{2}}-5m-84=0 \\
\end{align}\]
Now, by using the quadratic formula for the equation of type $a{{x}^{2}}+bx+c=0$ is given by:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So, by applying the above formula for the equation \[{{m}^{2}}-5m-84=0\], we get:
$\begin{align}
  & m=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 1 \right)\left( -84 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow m=\dfrac{5\pm \sqrt{25+336}}{2} \\
 & \Rightarrow m=\dfrac{5\pm \sqrt{361}}{2} \\
 & \Rightarrow m=\dfrac{5\pm 19}{2} \\
\end{align}$
So, we get the two values of m as:
$\begin{align}
  & m=\dfrac{5+19}{2} \\
 & \Rightarrow m=\dfrac{24}{2} \\
 & \Rightarrow m=12 \\
\end{align}$ and $\begin{align}
  & m=\dfrac{5-19}{2} \\
 & \Rightarrow m=\dfrac{-14}{2} \\
 & \Rightarrow m=-7 \\
\end{align}$
Now, we know that m represents the number of men participating in the tournament which can’t be negative.
So, we get the value of m as 12.
Hence, the option (c) is correct.

Note: Now, to solve these type of the questions we need to know some of the basics of combination to solve the question correctly. So, the basic formula of the combination is given by:
$^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
To find the factorial of the number n, multiply the number n with (n-1) till it reaches to 1. To understand let us find the factorial of 4.
$\begin{align}
  & 4!=4\times 3\times 2\times 1 \\
 & \Rightarrow 24 \\
\end{align}$