Question

In a chess tournament where the participants were to play one game with another, two players fell ill having played 3 games each. If the total number of games played is 84, the number of participants at the beginning was
 $ \begin{align}
  & A.\text{ 13} \\
 & B.\text{ 14} \\
 & C.\text{ 15} \\
 & D.\text{ 10} \\
\end{align} $

Answer 1

Hint: In this question two basic principles of counting are to be used. The first one is the multiplicative rule of counting and the other is the additive rule of counting. If a job can be done in m ways and others can be done by n ways and each job depends on each other then the total number of ways to do the job = $ \left( m \right)(n) $ . This is a multiplicative rule of counting.
Additive rule of counting: If a job can be done in m ways and other can be done by n ways and both job do not depends on each other that is they are mutually exclusive, then total number of way to do the job = $ \left( m \right)+(n) $ .This is additive rule of counting. In this question each participant plays with each other so each player plays with (n-1) players. In this question total game played is 84 and two players fell ill so, in order to calculate total game played we have to subtract the total game played when no one is ill minus total game played when two players are ill.

Complete step-by-step answer:
A chess game is played between two players, suppose A and B. It means if A plays with B or B plays with A both are the same event. Hence If there are n players and they play chess with each other, it means
First player plays with $ n-1 $ players
second player plays with $ n-2 $ players
 third player plays with $ n-3 $ players
And so-on
last player plays with $ 1 $ players
here we use the multiplicative rule of counting as one player played with n-1 other player.
Here number of games played by each player is mutually exclusive
Hence, total numbers of games are the sum of
 $ (n-1)+(n-2)+(n-3)+...+1 $
Here we use additive rule of counting
This is the sum of first (n-1) natural numbers. Suppose the sum is S
So, we can write
 $ S=1+2+3+...+n-1 $
As we know sum of first m natural numbers is given by
 $ \dfrac{m(m+1)}{2} $
So, using the above formula we can write
 $ \begin{align}
  & S=\dfrac{(n-1)(n-1+1)}{2} \\
 & \Rightarrow S=\dfrac{n(n-1)}{2} \\
\end{align} $
Now it is given from question that two players play 3 match each after that they ill
So, match played by them =6
As if they do not ill match will be played by them = $ (n-1)+(n-1)-1 $
Because each player played $ n-1 $ match, 1 is subtracted as in counting process the ill player played once but counted as 2 as in each $ (n-1) $
Now from question
Total match played is 84
So, we can write the above question in mathematical equation as
 $ \dfrac{n(n-1)}{2}-\left[ \left( n-1 \right)+\left( n-1 \right)-1 \right]+6=84-(a) $
See here
 $ \dfrac{n(n-1)}{2} $ are the total match if no player is ill
 $ 6 $ is the match played by the two-ill players.
 $ \left[ \left( n-1 \right)+\left( n-1 \right)-1 \right] $ will be the match played by the two people if they were not ill.
Now we have to solve equation $ (a) $ , taking LCM of denominator as 2 we can write
\[\begin{align}
  & \dfrac{n(n-1)-2\{2(n-1)-1\}+2(6)}{2}=84 \\
 & \Rightarrow {{n}^{2}}-n-4n+4+2+12=168 \\
 & \Rightarrow {{n}^{2}}-5n+18-168=0 \\
 & \Rightarrow {{n}^{2}}-5n-150=0 \\
\end{align}\]
Now here we split the mid-term in such a way that the product of them is a product of first term and last terms.
So, we can write the above quadratic equation as
 $ \begin{align}
  & {{n}^{2}}-15n+10n-150=0 \\
 & \Rightarrow n(n-15)+10(n-15)=0 \\
\end{align} $
Taking $ (n-15) $ as common we can write it as
 $ (n-15)(n+10)=0 $
So, this factor has two roots
 $ \begin{align}
  & n=15 \\
 & n=-10 \\
\end{align} $
As number of players cannot be negative
So, n=15
Hence the number of participants at the beginning was 15
So, option C is correct.


NOTE: In this question we have to select two participants out of n participants so here instead counting of total games we can use selection formula directly that is $ {}^{n}{{c}_{2}} $ .also we have $ {}^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} $ so $ {}^{n}{{c}_{2}}=\dfrac{n!}{2!\left( n-2 \right)!}=\dfrac{n(n-1)}{2} $
Also, for solving quadratic equation we can directly use the formula $ {{x}_{1,2}}=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $