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 (integer)
16
Solution
Let, $n$ be the total number of couples who participated in the tournament.
<br/><br/>According to the question, $2 \times{ }^n C_2 \times{ }^{n-2} C_2=840$
<br/><br/>$$
\begin{aligned}
& \Rightarrow{ }^n C_2 \times{ }^{n-2} C_2=420 \\\\
& \Rightarrow \frac{n !}{2 !(n-2) !} \times \frac{(n-2) !}{(n-4) ! 2 !}=420 \\\\
& \Rightarrow \frac{n(n-1)(n-2)(n-3)}{4}=420
\end{aligned}
$$
<br/><br/>Put $n=8$ satisfied the equation.
<br/><br/>So, $n=8$
<br/><br/>Hence, total number of players who participated<br/>in the tournament $=2 \times 8=16$
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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