Question

Two women and some men participated in a chess tournament in which every participant played two games with each of the other participants. If the number of games that the men played between themselves exceeds the number of games that the men played with the women by 66, then the number of men who participated in the tournament lies in the interval:
(a) [8, 9]
(b) [10, 12]
(c) (11, 13]
(d) (14, 17)

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 take the number of men who participated in the chess tournament to be x and also every man plays 2 games and out of x men only two play at a time. So the number of games played by men among themselves is $2\times {}^{x}{{C}_{2}}$.

Complete step-by-step answer:
A combination is a way to order or arrange a set or number of things uniquely. It is mentioned in the question that there are two women and some men. Now, let’s take the number of men who participated in the chess tournament to be x.
It is given that every man has played two games and out of x men only two play at a time. Using this information, we get,
Number of chess games played by men among themselves $=2\times {}^{x}{{C}_{2}}.......(1)$
Each woman can play two times and x men played with two women. Using this information, we get,
Number of chess games men played with women $=2\times 2x.......(2)$
It is given that the number of games that the men played between themselves exceeds the number of games that the men played with the women by 66. So taking this into account we subtract equation (2) from equation (1) and then equating it to 66.
$\Rightarrow 2\times {}^{x}{{C}_{2}}-2\times 2x=66.......(3)$
Now applying the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$ in equation (3), we get,

& \Rightarrow 2\times \dfrac{x!}{2!\,\times (x-2)!}-2\times 2x=66 \\\ & \Rightarrow 2\times \dfrac{x(x-1)(x-2)!}{2\,\times (x-2)!}-4x=66.........(4) \\\ \end{aligned}$$ Cancelling similar terms and solving for x we get, $$\begin{aligned} & \Rightarrow x(x-1)-4x=66 \\\ & \Rightarrow {{x}^{2}}-x-4x=66 \\\ & \Rightarrow {{x}^{2}}-5x-66=0 \\\ & \Rightarrow {{x}^{2}}-11x+6x-66=0 \\\ & \Rightarrow x(x-11)+6(x-11)=0 \\\ & \Rightarrow (x+6)\,(x-11)=0 \\\ & \Rightarrow x=-6,11 \\\ \end{aligned}$$ So the number of men who participated in the tournament is 11 because the number of men cannot be negative. Hence the number of men who participated in the tournament lies in the interval [10, 12]. And hence option (b) is the correct option. Note: Reading questions precisely which are related to combinations are very important so read it 3 to 4 times. And knowing this formula $${}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}$$ is important . We can make a mistake in solving equation (4) if we do not properly expand $${}^{x}{{C}_{2}}$$.