Question

In a class tournament, all participants were to play different games with one another, Two players fell ill after having played three games each. If the total number of games played in the tournament is equal to 84, the total number of participants in the beginning was equal to
(a) 10
(b) 15
(c) 12
(d) 14

Answer 1

Hint: We will solve this question with the help of combinations and hence we will use the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$. We will assume the players in the beginning to be n and at a time only two players play and hence the total number of games to be played will be ${}^{n}{{C}_{2}}$.

Complete step-by-step answer:
Let the players in the beginning be n.
Now the total number of games to be played $={}^{n}{{C}_{2}}........(1)$
And each player would have played n-1 games as it is mentioned in the question that all participants were to play different games with one another.
Now assuming that two players A and B fell ill.
So the total number of games of A and B will be n-1 plus n-1 and subtracting 1 as A and B play a game between them. So using this information we get,
Total number of games of A and B $=n-1+n-1-1=2n-3.......(2)$
But A and B fell ill after playing 3 matches,
So the remaining number of matches of A and B $=2n-3-6=2n-9.......(3)$
Now it is given that the total number of games played in the tournament is equal to 84. So subtracting equation (1) and equation (3) and equating it to 84 we get,
$\Rightarrow {}^{n}{{C}_{2}}-(2n-9)=84........(4)$
Now using the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$ in equation (4) we get,
$\Rightarrow \dfrac{n!}{2!(n-2)!}-(2n-9)=84........(5)$
Now expanding the factorials in equation (5) we get,
$\Rightarrow \dfrac{n(n-1)(n-2)!}{2(n-2)!}-(2n-9)=84........(6)$
Now cancelling similar terms in equation (6) and taking LCM and solving in equation (6) we get,

& \Rightarrow \dfrac{n(n-1)-2(2n-9)}{2}=84 \\\ & \Rightarrow n(n-1)-2(2n-9)=84\times 2........(7) \\\ \end{aligned}$$ Now simplifying the terms in equation (7) we get, $$\begin{aligned} & \Rightarrow {{n}^{2}}-n-4n+18=168 \\\ & \Rightarrow {{n}^{2}}-5n+18-168=0 \\\ & \Rightarrow {{n}^{2}}-5n-150=0.........(8) \\\ \end{aligned}$$ Now factoring in equation (8) we get, $$\begin{aligned} & \Rightarrow {{n}^{2}}-15n+10n-150=0 \\\ & \Rightarrow n(n-15)+10(n-15)=0 \\\ & \Rightarrow (n-15)(n+10)=0 \\\ & \Rightarrow n=15,-10 \\\ \end{aligned}$$ Hence the total number of participants in the beginning was equal to 15 as it cannot be -10. So the correct answer is option (b). Note: A combination is a way to order or arrange a set or number of things uniquely. And knowing this formula $${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$$ is important . We can make a mistake in solving equation (4) if we do not properly expand $${}^{n}{{C}_{2}}$$. Also we should know the expansion of factorials.