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 :

Let m be the number of men and 2 be the number of women.
The number of games played by the men between themselves is given by the combination formula:
²C₂ × 2 = m(m-1) × 2 / 2 = m(m-1)
The number of games played between the men and the women is given by:
mC₁ × ²C₁ × 2 = m × 2 × 2 = 4m
According to the problem, the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84. Therefore,
m(m-1) - 4m = 84
m² - m - 4m - 84 = 0
m² - 5m - 84 = 0
This is a quadratic equation. We can solve it by factoring:
(m - 12)(m + 7) = 0
This gives two possible solutions for m: m = 12 or m = -7.
Since the number of men cannot be negative, the value of m is 12.