Question

In a class tournament where the participants were to play one game with one another, two class players fell ill, having played 3 games each. If the total number of games played is 84, then the number of participants at the beginning was equal to:
(a) 10
(b) 15
(c) 12
(d) 14

Answer 1

Hint: Assume total number of participants to be n. Total number of games will be equal to $^{n}{{C}_{2}}$ if all the participants play with one another. Also, the total number of games played would be equal to the sum of total number of games played by $\left( n-2 \right)$ participants and three games played by each of the remaining two participants.

Complete step-by-step answer:
We have to find the total number of participants, so let us assume the total number of participants be n.
According to the question, each participant plays with other participants.
So, the total number of games played will be equal to the total number of ways of selecting two participants out of ’n’ participants.
So the total number of games = $^{n}{{C}_{2}}$ .
But according to the question, two of the participants won’t play games with all other participants but they played 3 games each and then fell ill.
So, let us calculate all the games played by $\left( n-2 \right)$ participants, which will be equal to the number of ways of selecting participants out of $\left( n-2 \right)$ participants.
So, total no. of games played by $\left( n-2 \right)$ participants = $^{\left( n-2 \right)}{{C}_{2}}$
Now, two of the participants played 3 games each.
So, total number of games= $^{\left( n-2 \right)}{{C}_{3}}+2\left( 3 \right)$
And according to the question, the total number of games played = 84.
So, $^{\left( n-2 \right)}{{C}_{2}}+6=84$
\[\begin{align}
  & {{\Rightarrow }^{\left( n-2 \right)}}{{C}_{2}}=84-6 \\
 & =78 \\
 & \\
\end{align}\]
\[{{\Rightarrow }^{\left( n-2 \right)}}{{C}_{2}}=84-6\]
\[{{\Rightarrow }^{\left( n-2 \right)}}{{C}_{2}}=78\]
As, $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
$\Rightarrow \dfrac{\left( n-2 \right)!}{2!\left( n-2-2 \right)!}=78$
$\Rightarrow \dfrac{\left( n-2 \right)!}{2!\left( n-4 \right)!}=78$
Now, we know that $\left( n-2 \right)!=\left( n-2 \right)\times \left( n-2-1 \right)\times \left( n-2-2 \right)!=\left( n-2 \right)\times \left( n-3 \right)\times \left( n-4 \right)!$
So, equation will become:
$\dfrac{\left( n-2 \right)\times \left( n-3 \right)\times \left( n-4 \right)!}{2\times \left( n-4 \right)!}=78$
Dividing both numerator and denominator of LHS by (n-4)!, we will get:
$\dfrac{\left( n-2 \right)\left( n-3 \right)}{2}=78$
Multiplying both sides by 2, we will get,
$\begin{align}
  & \left( n-2 \right)\left( n-3 \right)=156 \\
 & \Rightarrow {{n}^{2}}-3n-2n+6=156 \\
\end{align}$
Taking =0 all the terms to LHS, we will get,
\[\begin{align}
  & \Rightarrow {{n}^{2}}-5n+6-156=0 \\
 & \Rightarrow {{n}^{2}}-5n-150=0 \\
\end{align}\]
Now, we have got a quadratic equation to solve this quadratic equation let us factorize this equation by splitting its middle term: -
${{n}^{2}}-15n+10n-150=0$
Taking ‘n’ common from first two terms and ’10; common from last two terms, we will get,
$\begin{align}
  & n\left( n-15 \right)+10\left( n-15 \right)=0 \\
 & \left( n+10 \right)\left( n-15 \right)=0 \\
 & \Rightarrow n=-10\,\,or\,n=15 \\
\end{align}$

But the number of participants can’t be negative. So, n=15.
Hence, the required number of participants to the beginning is 15 and option (b) is the correct answer.

Note: The quadratic equation in n, i.e. \[{{n}^{2}}-5n-150=0\] , can be solved by the quadratic formula. We know, for the quadratic equation $a{{x}^{2}}+bx+c=0$ , the roots are given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . So, the roots of the quadratic equation \[{{n}^{2}}-5n-150=0\] are given as $n=\dfrac{5\pm \sqrt{25+600}}{2}$ .
$\Rightarrow n=\dfrac{5\pm \sqrt{625}}{2}$
$\Rightarrow n=\dfrac{5\pm 25}{2}$
$\Rightarrow n=\dfrac{30}{2},\dfrac{-20}{2}$
$\Rightarrow n=15,-10$.
Since n is the number of participants, it cannot be negative. Hence, n = 15.