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 numbers of persons, who participated in the tournament, is _______.

Some couples participated in a mixed doubles badminton tournament with no couple playing in a match. Total matches = 840. Find total persons.

Let there be $$n$$ couples ($$2n$$ persons total). In mixed doubles, each team has one male and one female. For one match, choose 2 males in $$\binom{n}{2}$$ ways, then choose 2 females who are not the wives of the chosen males in $$\binom{n-2}{2}$$ ways, and pair them into two teams in 2 ways. Since each pairing corresponds to one distinct match, the total number of matches is $$\binom{n}{2}\binom{n-2}{2} \times 2 = 840$$. Substituting the binomial coefficients gives $$\dfrac{n(n-1)}{2} \times \dfrac{(n-2)(n-3)}{2} \times 2 = 840$$, which simplifies to $$\dfrac{n(n-1)(n-2)(n-3)}{2} = 840$$ and hence $$n(n-1)(n-2)(n-3) = 1680$$. Testing $$n = 8$$: $$8 \times 7 \times 6 \times 5 = 1680$$ $$\checkmark$$, so $$2n = 16$$.

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