Some couples participated in a mixed doubles badminton tournament.

Question

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

Accepted Answer

Let, $n$ be the total number of couples who participated in the tournament.

According to the question, $2	imes {}^{n}{C}_{2}	imes {}^{n-2}{C}_{2}=840$

$\begin{gathered} \Rightarrow {}^{n}{C}_{2}	imes {}^{n-2}{C}_{2}=420 \ \Rightarrow \frac{n!}{2!(n-2)!}	imes \frac{(n-2)!}{(n-4)!2!}=420 \ \Rightarrow \frac{n(n-1)(n-2)(n-3)}{4}=420 \end{gathered}$

Put $n=8$ satisfied the equation.

So, $n=8$

Hence, total number of players who participated
in the tournament $=2	imes 8=16$

Solutions · 1