JEE MAIN - Mathematics (2023 - 10th April Morning Shift - No. 22)
Some couples participated in a mixed doubles badminton tournament. If the number of matches played, so that no couple played in a match, is 840, then the total number of persons, who participated in the tournament, is ___________.
Answer
16
Explanation
Let, $n$ be the total number of couples who participated in the tournament.
According to the question, $2 \times{ }^n C_2 \times{ }^{n-2} C_2=840$
$$ \begin{aligned} & \Rightarrow{ }^n C_2 \times{ }^{n-2} C_2=420 \\\\ & \Rightarrow \frac{n !}{2 !(n-2) !} \times \frac{(n-2) !}{(n-4) ! 2 !}=420 \\\\ & \Rightarrow \frac{n(n-1)(n-2)(n-3)}{4}=420 \end{aligned} $$
Put $n=8$ satisfied the equation.
So, $n=8$
Hence, total number of players who participated
in the tournament $=2 \times 8=16$
According to the question, $2 \times{ }^n C_2 \times{ }^{n-2} C_2=840$
$$ \begin{aligned} & \Rightarrow{ }^n C_2 \times{ }^{n-2} C_2=420 \\\\ & \Rightarrow \frac{n !}{2 !(n-2) !} \times \frac{(n-2) !}{(n-4) ! 2 !}=420 \\\\ & \Rightarrow \frac{n(n-1)(n-2)(n-3)}{4}=420 \end{aligned} $$
Put $n=8$ satisfied the equation.
So, $n=8$
Hence, total number of players who participated
in the tournament $=2 \times 8=16$
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