Permutations and Combinations 2 Question 4
4. 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 (2019 Main, 12 Jan II)
(a) 12
(b) 11
(c) 9
(d) 7
Show Answer
Answer:
Correct Answer: 4. (a)
Solution:
- Since, there are m-men and 2-women and each participant plays two games with every other participant.
$\therefore$ Number of games played by the men between themselves $=2 \times{ }^{m} C _2$
and the number of games played between the men and the women $\Rightarrow \times{ }^{m} C _1 \times{ }^{2} C _1$
Now, according to the question,
$$ \begin{aligned} & { } _{m !}{ }^{m} C _2=2{ }^{m} C _1{ }^{2} C _1+84 \\ & \Rightarrow \quad \frac{m !}{2 !(m-2) !}=m \times 2+42 \\ & \Rightarrow \quad m(m-1)=4 m+84 \\ & \Rightarrow \quad m^{2}-m=4 m+84 \end{aligned} $$
$$ \begin{array}{rlr} \Rightarrow & m^{2}-5 m-84=0 \\ \Rightarrow \quad & m^{2}-12 m+7 m-84 & =0 \\ \Rightarrow & m(m-12)+7(m-12) & =0 \\ \Rightarrow & m & =12 \end{array} $$
$[\because m>0]$
